QBManager

Some Basic Concepts of Chemistry

228699 Calculate the amount of hydrogen gas required in order to produce $100 g$ of ammonia by the reaction of $N_{2} (g)$ and $H_{2} (g)$ gases.

1

35.29 g

2

17.65 g

3

28.11 g

4

34 g

AP

Explanation:

Given, ${NH}_{3}$ Produce $100 gm$ . $N_{2} (g) + 3 H_{2} (g) ⟶ 2 {NH}_{3}$ So, the number of moles of
$\begin{aligned} {NH}_{3} & = \frac{Weight}{Molecular weight} \\ n_{{NH}_{3}} & = \frac{100}{17} \\ n_{{NH}_{3}} & = 5.88 mole. \end{aligned}$
So, the amount of hydrogen gas
Required is $= 3 \times 5.88 = 17.65 gm$ .

AP EAPCET 24.08.2021 Shift-II

Some Basic Concepts of Chemistry

228700 When 20 g of ${CaCO}_{3}$ is treated with $20 g$ of $HCl$ the mass of ${CO}_{2}$ formed would be

1

10 g

2

8.8 g

3

22.2 g

4

20 g

Ans.

Explanation:

$\begin{array}{ccc} {CaCO}_{3} (s) & + & 2 HCl (aq) & \to & {CaCl}_{2} (aq) & + & H_{2} O (l) & + & {CO}_{2} (g) \\ 1 mol & 2 mol & 1 mol \\ 100 g & 73 g & 44 g \end{array}$
Let ${CaCO}_{3} (s)$ be completely consumed in the reaction. $∴ 100 g {CaCO}_{3}$ give $44 g {CO}_{2}$
$∴ 20 g {CaCO}_{3}$ will give $\frac{44}{100} \times 20 g {CO}_{2} = 8.8 g {CO}_{2}$

AP EAPCET 19-08-2021 Shift-I

Some Basic Concepts of Chemistry

228701 The complete combustion of one mole of benzene produces grams of carbon dioxide.

1 164

2 220

3 264

4 308

Explanation:

Combustion reaction of benzene to carbon dioxide is-
$C_{6} H_{6} + \frac{15}{2} O_{2} (g) ⟶ 6 {CO}_{2} (g) + 3 H_{2} O (g)$
Find grams of carbon dioxide.
$\begin{aligned} ∴ mole = \frac{weight}{moleculor weight} \\ 6 = \frac{weight}{44} \\ Weight = 44 \times 6 = 264 gm \end{aligned}$
So, 1 mole of benzene produces 264 gm of carbon dioxide.

AP EAPCET 20.08.2021 Shift-II

Some Basic Concepts of Chemistry

228705 The minimum amount of $O_{2} (g)$ consumed per gram of reactant is for the reaction (Given atomic mass : $F e = 56, O = 16, Mg = 24, P = 31$ , ) $C = 12, H = 1$ )

1

C_{3} H_{8}

(g)

+ 5 O_{2}

(g)

\to 3 {CO}_{2}

(g)

+ 4 H_{2} O (l)

2

P_{4}

(s)

+ 5 O_{2}

(g)

\to P_{4} O_{10}

(s)

3

4 Fe (s) + 3 O_{2}

(g)

\to 2 {Fe}_{2} O_{3}

(s)

4

2 Mg (s) + O_{2}

(g)

\to 2 MgO

(s)

Explanation:

Option (a)- $C_{3} H_{8} (g) + 5 O_{2} (g) \to 3 {XO}_{2} (g) + 4 H_{2} O (l)$
$(1 \times 36 gm) + (1 \times 8 gm) = 5 \times 32 gm$
$36 + 8 = 160$
$1 gm = \frac{160}{44} = 3.63 gm$
Option (b) $- P_{4} (S) + 5 O_{2} (g) \to P_{4} O_{10} (s)$
$4 \times 31 gm = 5 \times 32 gm$
$1 gm = \frac{5 \times 32}{4 \times 31} = 1.29 gm$
Each $1 g$ of $p$ requires $1.29 g$ of $O_{2}$ .
Option (c)- $4 Fe (s) + 3 O_{2} (g) \to 2 {Fe}_{2} O_{3} (s)$ :
$4 \times 56 gm = 3 \times 32 gm$
$1 gm = \frac{3 \times 32}{4 \times 56} = 0.428 gm$
Each $1 g$ of Fe requires $0.428 g$ of $O_{2}$
Option (d) $- 2 Mg (s) + O_{2} (g) \to 2 MgO (s)$
$2 \times 24 gm = 1 \times 32 gm$
$1 gm = \frac{1 \times 32}{2 \times 24} = 0.67 gm$
Each $1 g$ of $Mg$ requires 0.67 of $O_{2}$ .

[JEE Main 2019

Some Basic Concepts of Chemistry

228706 $25 g$ of an unknown hydrocarbon upon burning produces 88 g of ${CO}_{2}$ and $9 g$ of $H_{2} O$ . This unknown hydrocarbon contains

1

20 g

of carbon and

5 g

of hydrogen

2

22 g

of carbon and

3 g

of hydrogen

3

24 g

of carbon and

1 g

of hydrogen

4

18 g

of carbon and

7 g

of hydrogen

Explanation:

Let the hydrocarbon is $C_{x} H_{y}$ .
$C_{x} H_{y} + O_{2} ⟶ {xCO}_{2} + \frac{y}{2} H_{2} O$
The weight of Carbon $= \frac{88}{44} \times 12$
$= 24 g$
The weight of hydrogen $= \frac{9}{18} \times 1$
$\begin{aligned} = 24 g \\ C_{x} H_{y} + (x + \frac{y}{4}) O_{2} \to {xCO}_{2} + \frac{y}{2} H_{2} O \\ C = x \times \frac{25}{M} = 2 \\ H = \frac{y}{2} \times \frac{25}{M} = 1 \\ \Rightarrow 24 : 1 ratio by mass \end{aligned}$

[JEE Main 2019

Some Basic Concepts of Chemistry

228699 Calculate the amount of hydrogen gas required in order to produce $100 g$ of ammonia by the reaction of $N_{2} (g)$ and $H_{2} (g)$ gases.

1

35.29 g

2

17.65 g

3

28.11 g

4

34 g

AP

Explanation:

Given, ${NH}_{3}$ Produce $100 gm$ . $N_{2} (g) + 3 H_{2} (g) ⟶ 2 {NH}_{3}$ So, the number of moles of
$\begin{aligned} {NH}_{3} & = \frac{Weight}{Molecular weight} \\ n_{{NH}_{3}} & = \frac{100}{17} \\ n_{{NH}_{3}} & = 5.88 mole. \end{aligned}$
So, the amount of hydrogen gas
Required is $= 3 \times 5.88 = 17.65 gm$ .

AP EAPCET 24.08.2021 Shift-II

Some Basic Concepts of Chemistry

228700 When 20 g of ${CaCO}_{3}$ is treated with $20 g$ of $HCl$ the mass of ${CO}_{2}$ formed would be

1

10 g

2

8.8 g

3

22.2 g

4

20 g

Ans.

Explanation:

$\begin{array}{ccc} {CaCO}_{3} (s) & + & 2 HCl (aq) & \to & {CaCl}_{2} (aq) & + & H_{2} O (l) & + & {CO}_{2} (g) \\ 1 mol & 2 mol & 1 mol \\ 100 g & 73 g & 44 g \end{array}$
Let ${CaCO}_{3} (s)$ be completely consumed in the reaction. $∴ 100 g {CaCO}_{3}$ give $44 g {CO}_{2}$
$∴ 20 g {CaCO}_{3}$ will give $\frac{44}{100} \times 20 g {CO}_{2} = 8.8 g {CO}_{2}$

AP EAPCET 19-08-2021 Shift-I

Some Basic Concepts of Chemistry

228701 The complete combustion of one mole of benzene produces grams of carbon dioxide.

1 164

2 220

3 264

4 308

Explanation:

Combustion reaction of benzene to carbon dioxide is-
$C_{6} H_{6} + \frac{15}{2} O_{2} (g) ⟶ 6 {CO}_{2} (g) + 3 H_{2} O (g)$
Find grams of carbon dioxide.
$\begin{aligned} ∴ mole = \frac{weight}{moleculor weight} \\ 6 = \frac{weight}{44} \\ Weight = 44 \times 6 = 264 gm \end{aligned}$
So, 1 mole of benzene produces 264 gm of carbon dioxide.

AP EAPCET 20.08.2021 Shift-II

Some Basic Concepts of Chemistry

228705 The minimum amount of $O_{2} (g)$ consumed per gram of reactant is for the reaction (Given atomic mass : $F e = 56, O = 16, Mg = 24, P = 31$ , ) $C = 12, H = 1$ )

1

C_{3} H_{8}

(g)

+ 5 O_{2}

(g)

\to 3 {CO}_{2}

(g)

+ 4 H_{2} O (l)

2

P_{4}

(s)

+ 5 O_{2}

(g)

\to P_{4} O_{10}

(s)

3

4 Fe (s) + 3 O_{2}

(g)

\to 2 {Fe}_{2} O_{3}

(s)

4

2 Mg (s) + O_{2}

(g)

\to 2 MgO

(s)

Explanation:

Option (a)- $C_{3} H_{8} (g) + 5 O_{2} (g) \to 3 {XO}_{2} (g) + 4 H_{2} O (l)$
$(1 \times 36 gm) + (1 \times 8 gm) = 5 \times 32 gm$
$36 + 8 = 160$
$1 gm = \frac{160}{44} = 3.63 gm$
Option (b) $- P_{4} (S) + 5 O_{2} (g) \to P_{4} O_{10} (s)$
$4 \times 31 gm = 5 \times 32 gm$
$1 gm = \frac{5 \times 32}{4 \times 31} = 1.29 gm$
Each $1 g$ of $p$ requires $1.29 g$ of $O_{2}$ .
Option (c)- $4 Fe (s) + 3 O_{2} (g) \to 2 {Fe}_{2} O_{3} (s)$ :
$4 \times 56 gm = 3 \times 32 gm$
$1 gm = \frac{3 \times 32}{4 \times 56} = 0.428 gm$
Each $1 g$ of Fe requires $0.428 g$ of $O_{2}$
Option (d) $- 2 Mg (s) + O_{2} (g) \to 2 MgO (s)$
$2 \times 24 gm = 1 \times 32 gm$
$1 gm = \frac{1 \times 32}{2 \times 24} = 0.67 gm$
Each $1 g$ of $Mg$ requires 0.67 of $O_{2}$ .

[JEE Main 2019

Some Basic Concepts of Chemistry

228706 $25 g$ of an unknown hydrocarbon upon burning produces 88 g of ${CO}_{2}$ and $9 g$ of $H_{2} O$ . This unknown hydrocarbon contains

1

20 g

of carbon and

5 g

of hydrogen

2

22 g

of carbon and

3 g

of hydrogen

3

24 g

of carbon and

1 g

of hydrogen

4

18 g

of carbon and

7 g

of hydrogen

Explanation:

Let the hydrocarbon is $C_{x} H_{y}$ .
$C_{x} H_{y} + O_{2} ⟶ {xCO}_{2} + \frac{y}{2} H_{2} O$
The weight of Carbon $= \frac{88}{44} \times 12$
$= 24 g$
The weight of hydrogen $= \frac{9}{18} \times 1$
$\begin{aligned} = 24 g \\ C_{x} H_{y} + (x + \frac{y}{4}) O_{2} \to {xCO}_{2} + \frac{y}{2} H_{2} O \\ C = x \times \frac{25}{M} = 2 \\ H = \frac{y}{2} \times \frac{25}{M} = 1 \\ \Rightarrow 24 : 1 ratio by mass \end{aligned}$

[JEE Main 2019

Some Basic Concepts of Chemistry

228699 Calculate the amount of hydrogen gas required in order to produce $100 g$ of ammonia by the reaction of $N_{2} (g)$ and $H_{2} (g)$ gases.

1

35.29 g

2

17.65 g

3

28.11 g

4

34 g

AP

Explanation:

Given, ${NH}_{3}$ Produce $100 gm$ . $N_{2} (g) + 3 H_{2} (g) ⟶ 2 {NH}_{3}$ So, the number of moles of
$\begin{aligned} {NH}_{3} & = \frac{Weight}{Molecular weight} \\ n_{{NH}_{3}} & = \frac{100}{17} \\ n_{{NH}_{3}} & = 5.88 mole. \end{aligned}$
So, the amount of hydrogen gas
Required is $= 3 \times 5.88 = 17.65 gm$ .

AP EAPCET 24.08.2021 Shift-II

Some Basic Concepts of Chemistry

228700 When 20 g of ${CaCO}_{3}$ is treated with $20 g$ of $HCl$ the mass of ${CO}_{2}$ formed would be

1

10 g

2

8.8 g

3

22.2 g

4

20 g

Ans.

Explanation:

$\begin{array}{ccc} {CaCO}_{3} (s) & + & 2 HCl (aq) & \to & {CaCl}_{2} (aq) & + & H_{2} O (l) & + & {CO}_{2} (g) \\ 1 mol & 2 mol & 1 mol \\ 100 g & 73 g & 44 g \end{array}$
Let ${CaCO}_{3} (s)$ be completely consumed in the reaction. $∴ 100 g {CaCO}_{3}$ give $44 g {CO}_{2}$
$∴ 20 g {CaCO}_{3}$ will give $\frac{44}{100} \times 20 g {CO}_{2} = 8.8 g {CO}_{2}$

AP EAPCET 19-08-2021 Shift-I

Some Basic Concepts of Chemistry

228701 The complete combustion of one mole of benzene produces grams of carbon dioxide.

1 164

2 220

3 264

4 308

Explanation:

Combustion reaction of benzene to carbon dioxide is-
$C_{6} H_{6} + \frac{15}{2} O_{2} (g) ⟶ 6 {CO}_{2} (g) + 3 H_{2} O (g)$
Find grams of carbon dioxide.
$\begin{aligned} ∴ mole = \frac{weight}{moleculor weight} \\ 6 = \frac{weight}{44} \\ Weight = 44 \times 6 = 264 gm \end{aligned}$
So, 1 mole of benzene produces 264 gm of carbon dioxide.

AP EAPCET 20.08.2021 Shift-II

Some Basic Concepts of Chemistry

228705 The minimum amount of $O_{2} (g)$ consumed per gram of reactant is for the reaction (Given atomic mass : $F e = 56, O = 16, Mg = 24, P = 31$ , ) $C = 12, H = 1$ )

1

C_{3} H_{8}

(g)

+ 5 O_{2}

(g)

\to 3 {CO}_{2}

(g)

+ 4 H_{2} O (l)

2

P_{4}

(s)

+ 5 O_{2}

(g)

\to P_{4} O_{10}

(s)

3

4 Fe (s) + 3 O_{2}

(g)

\to 2 {Fe}_{2} O_{3}

(s)

4

2 Mg (s) + O_{2}

(g)

\to 2 MgO

(s)

Explanation:

Option (a)- $C_{3} H_{8} (g) + 5 O_{2} (g) \to 3 {XO}_{2} (g) + 4 H_{2} O (l)$
$(1 \times 36 gm) + (1 \times 8 gm) = 5 \times 32 gm$
$36 + 8 = 160$
$1 gm = \frac{160}{44} = 3.63 gm$
Option (b) $- P_{4} (S) + 5 O_{2} (g) \to P_{4} O_{10} (s)$
$4 \times 31 gm = 5 \times 32 gm$
$1 gm = \frac{5 \times 32}{4 \times 31} = 1.29 gm$
Each $1 g$ of $p$ requires $1.29 g$ of $O_{2}$ .
Option (c)- $4 Fe (s) + 3 O_{2} (g) \to 2 {Fe}_{2} O_{3} (s)$ :
$4 \times 56 gm = 3 \times 32 gm$
$1 gm = \frac{3 \times 32}{4 \times 56} = 0.428 gm$
Each $1 g$ of Fe requires $0.428 g$ of $O_{2}$
Option (d) $- 2 Mg (s) + O_{2} (g) \to 2 MgO (s)$
$2 \times 24 gm = 1 \times 32 gm$
$1 gm = \frac{1 \times 32}{2 \times 24} = 0.67 gm$
Each $1 g$ of $Mg$ requires 0.67 of $O_{2}$ .

[JEE Main 2019

Some Basic Concepts of Chemistry

228706 $25 g$ of an unknown hydrocarbon upon burning produces 88 g of ${CO}_{2}$ and $9 g$ of $H_{2} O$ . This unknown hydrocarbon contains

1

20 g

of carbon and

5 g

of hydrogen

2

22 g

of carbon and

3 g

of hydrogen

3

24 g

of carbon and

1 g

of hydrogen

4

18 g

of carbon and

7 g

of hydrogen

Explanation:

Let the hydrocarbon is $C_{x} H_{y}$ .
$C_{x} H_{y} + O_{2} ⟶ {xCO}_{2} + \frac{y}{2} H_{2} O$
The weight of Carbon $= \frac{88}{44} \times 12$
$= 24 g$
The weight of hydrogen $= \frac{9}{18} \times 1$
$\begin{aligned} = 24 g \\ C_{x} H_{y} + (x + \frac{y}{4}) O_{2} \to {xCO}_{2} + \frac{y}{2} H_{2} O \\ C = x \times \frac{25}{M} = 2 \\ H = \frac{y}{2} \times \frac{25}{M} = 1 \\ \Rightarrow 24 : 1 ratio by mass \end{aligned}$

[JEE Main 2019

Some Basic Concepts of Chemistry

228699 Calculate the amount of hydrogen gas required in order to produce $100 g$ of ammonia by the reaction of $N_{2} (g)$ and $H_{2} (g)$ gases.

1

35.29 g

2

17.65 g

3

28.11 g

4

34 g

AP

Explanation:

Given, ${NH}_{3}$ Produce $100 gm$ . $N_{2} (g) + 3 H_{2} (g) ⟶ 2 {NH}_{3}$ So, the number of moles of
$\begin{aligned} {NH}_{3} & = \frac{Weight}{Molecular weight} \\ n_{{NH}_{3}} & = \frac{100}{17} \\ n_{{NH}_{3}} & = 5.88 mole. \end{aligned}$
So, the amount of hydrogen gas
Required is $= 3 \times 5.88 = 17.65 gm$ .

AP EAPCET 24.08.2021 Shift-II

Some Basic Concepts of Chemistry

228700 When 20 g of ${CaCO}_{3}$ is treated with $20 g$ of $HCl$ the mass of ${CO}_{2}$ formed would be

1

10 g

2

8.8 g

3

22.2 g

4

20 g

Ans.

Explanation:

$\begin{array}{ccc} {CaCO}_{3} (s) & + & 2 HCl (aq) & \to & {CaCl}_{2} (aq) & + & H_{2} O (l) & + & {CO}_{2} (g) \\ 1 mol & 2 mol & 1 mol \\ 100 g & 73 g & 44 g \end{array}$
Let ${CaCO}_{3} (s)$ be completely consumed in the reaction. $∴ 100 g {CaCO}_{3}$ give $44 g {CO}_{2}$
$∴ 20 g {CaCO}_{3}$ will give $\frac{44}{100} \times 20 g {CO}_{2} = 8.8 g {CO}_{2}$

AP EAPCET 19-08-2021 Shift-I

Some Basic Concepts of Chemistry

228701 The complete combustion of one mole of benzene produces grams of carbon dioxide.

1 164

2 220

3 264

4 308

Explanation:

Combustion reaction of benzene to carbon dioxide is-
$C_{6} H_{6} + \frac{15}{2} O_{2} (g) ⟶ 6 {CO}_{2} (g) + 3 H_{2} O (g)$
Find grams of carbon dioxide.
$\begin{aligned} ∴ mole = \frac{weight}{moleculor weight} \\ 6 = \frac{weight}{44} \\ Weight = 44 \times 6 = 264 gm \end{aligned}$
So, 1 mole of benzene produces 264 gm of carbon dioxide.

AP EAPCET 20.08.2021 Shift-II

Some Basic Concepts of Chemistry

228705 The minimum amount of $O_{2} (g)$ consumed per gram of reactant is for the reaction (Given atomic mass : $F e = 56, O = 16, Mg = 24, P = 31$ , ) $C = 12, H = 1$ )

1

C_{3} H_{8}

(g)

+ 5 O_{2}

(g)

\to 3 {CO}_{2}

(g)

+ 4 H_{2} O (l)

2

P_{4}

(s)

+ 5 O_{2}

(g)

\to P_{4} O_{10}

(s)

3

4 Fe (s) + 3 O_{2}

(g)

\to 2 {Fe}_{2} O_{3}

(s)

4

2 Mg (s) + O_{2}

(g)

\to 2 MgO

(s)

Explanation:

Option (a)- $C_{3} H_{8} (g) + 5 O_{2} (g) \to 3 {XO}_{2} (g) + 4 H_{2} O (l)$
$(1 \times 36 gm) + (1 \times 8 gm) = 5 \times 32 gm$
$36 + 8 = 160$
$1 gm = \frac{160}{44} = 3.63 gm$
Option (b) $- P_{4} (S) + 5 O_{2} (g) \to P_{4} O_{10} (s)$
$4 \times 31 gm = 5 \times 32 gm$
$1 gm = \frac{5 \times 32}{4 \times 31} = 1.29 gm$
Each $1 g$ of $p$ requires $1.29 g$ of $O_{2}$ .
Option (c)- $4 Fe (s) + 3 O_{2} (g) \to 2 {Fe}_{2} O_{3} (s)$ :
$4 \times 56 gm = 3 \times 32 gm$
$1 gm = \frac{3 \times 32}{4 \times 56} = 0.428 gm$
Each $1 g$ of Fe requires $0.428 g$ of $O_{2}$
Option (d) $- 2 Mg (s) + O_{2} (g) \to 2 MgO (s)$
$2 \times 24 gm = 1 \times 32 gm$
$1 gm = \frac{1 \times 32}{2 \times 24} = 0.67 gm$
Each $1 g$ of $Mg$ requires 0.67 of $O_{2}$ .

[JEE Main 2019

Some Basic Concepts of Chemistry

228706 $25 g$ of an unknown hydrocarbon upon burning produces 88 g of ${CO}_{2}$ and $9 g$ of $H_{2} O$ . This unknown hydrocarbon contains

1

20 g

of carbon and

5 g

of hydrogen

2

22 g

of carbon and

3 g

of hydrogen

3

24 g

of carbon and

1 g

of hydrogen

4

18 g

of carbon and

7 g

of hydrogen

Explanation:

Let the hydrocarbon is $C_{x} H_{y}$ .
$C_{x} H_{y} + O_{2} ⟶ {xCO}_{2} + \frac{y}{2} H_{2} O$
The weight of Carbon $= \frac{88}{44} \times 12$
$= 24 g$
The weight of hydrogen $= \frac{9}{18} \times 1$
$\begin{aligned} = 24 g \\ C_{x} H_{y} + (x + \frac{y}{4}) O_{2} \to {xCO}_{2} + \frac{y}{2} H_{2} O \\ C = x \times \frac{25}{M} = 2 \\ H = \frac{y}{2} \times \frac{25}{M} = 1 \\ \Rightarrow 24 : 1 ratio by mass \end{aligned}$

[JEE Main 2019

Some Basic Concepts of Chemistry

228699 Calculate the amount of hydrogen gas required in order to produce $100 g$ of ammonia by the reaction of $N_{2} (g)$ and $H_{2} (g)$ gases.

1

35.29 g

2

17.65 g

3

28.11 g

4

34 g

AP

Explanation:

Given, ${NH}_{3}$ Produce $100 gm$ . $N_{2} (g) + 3 H_{2} (g) ⟶ 2 {NH}_{3}$ So, the number of moles of
$\begin{aligned} {NH}_{3} & = \frac{Weight}{Molecular weight} \\ n_{{NH}_{3}} & = \frac{100}{17} \\ n_{{NH}_{3}} & = 5.88 mole. \end{aligned}$
So, the amount of hydrogen gas
Required is $= 3 \times 5.88 = 17.65 gm$ .

AP EAPCET 24.08.2021 Shift-II

Some Basic Concepts of Chemistry

228700 When 20 g of ${CaCO}_{3}$ is treated with $20 g$ of $HCl$ the mass of ${CO}_{2}$ formed would be

1

10 g

2

8.8 g

3

22.2 g

4

20 g

Ans.

Explanation:

$\begin{array}{ccc} {CaCO}_{3} (s) & + & 2 HCl (aq) & \to & {CaCl}_{2} (aq) & + & H_{2} O (l) & + & {CO}_{2} (g) \\ 1 mol & 2 mol & 1 mol \\ 100 g & 73 g & 44 g \end{array}$
Let ${CaCO}_{3} (s)$ be completely consumed in the reaction. $∴ 100 g {CaCO}_{3}$ give $44 g {CO}_{2}$
$∴ 20 g {CaCO}_{3}$ will give $\frac{44}{100} \times 20 g {CO}_{2} = 8.8 g {CO}_{2}$

AP EAPCET 19-08-2021 Shift-I

Some Basic Concepts of Chemistry

228701 The complete combustion of one mole of benzene produces grams of carbon dioxide.

1 164

2 220

3 264

4 308

Explanation:

Combustion reaction of benzene to carbon dioxide is-
$C_{6} H_{6} + \frac{15}{2} O_{2} (g) ⟶ 6 {CO}_{2} (g) + 3 H_{2} O (g)$
Find grams of carbon dioxide.
$\begin{aligned} ∴ mole = \frac{weight}{moleculor weight} \\ 6 = \frac{weight}{44} \\ Weight = 44 \times 6 = 264 gm \end{aligned}$
So, 1 mole of benzene produces 264 gm of carbon dioxide.

AP EAPCET 20.08.2021 Shift-II

Some Basic Concepts of Chemistry

228705 The minimum amount of $O_{2} (g)$ consumed per gram of reactant is for the reaction (Given atomic mass : $F e = 56, O = 16, Mg = 24, P = 31$ , ) $C = 12, H = 1$ )

1

C_{3} H_{8}

(g)

+ 5 O_{2}

(g)

\to 3 {CO}_{2}

(g)

+ 4 H_{2} O (l)

2

P_{4}

(s)

+ 5 O_{2}

(g)

\to P_{4} O_{10}

(s)

3

4 Fe (s) + 3 O_{2}

(g)

\to 2 {Fe}_{2} O_{3}

(s)

4

2 Mg (s) + O_{2}

(g)

\to 2 MgO

(s)

Explanation:

Option (a)- $C_{3} H_{8} (g) + 5 O_{2} (g) \to 3 {XO}_{2} (g) + 4 H_{2} O (l)$
$(1 \times 36 gm) + (1 \times 8 gm) = 5 \times 32 gm$
$36 + 8 = 160$
$1 gm = \frac{160}{44} = 3.63 gm$
Option (b) $- P_{4} (S) + 5 O_{2} (g) \to P_{4} O_{10} (s)$
$4 \times 31 gm = 5 \times 32 gm$
$1 gm = \frac{5 \times 32}{4 \times 31} = 1.29 gm$
Each $1 g$ of $p$ requires $1.29 g$ of $O_{2}$ .
Option (c)- $4 Fe (s) + 3 O_{2} (g) \to 2 {Fe}_{2} O_{3} (s)$ :
$4 \times 56 gm = 3 \times 32 gm$
$1 gm = \frac{3 \times 32}{4 \times 56} = 0.428 gm$
Each $1 g$ of Fe requires $0.428 g$ of $O_{2}$
Option (d) $- 2 Mg (s) + O_{2} (g) \to 2 MgO (s)$
$2 \times 24 gm = 1 \times 32 gm$
$1 gm = \frac{1 \times 32}{2 \times 24} = 0.67 gm$
Each $1 g$ of $Mg$ requires 0.67 of $O_{2}$ .

[JEE Main 2019

Some Basic Concepts of Chemistry

228706 $25 g$ of an unknown hydrocarbon upon burning produces 88 g of ${CO}_{2}$ and $9 g$ of $H_{2} O$ . This unknown hydrocarbon contains

1

20 g

of carbon and

5 g

of hydrogen

2

22 g

of carbon and

3 g

of hydrogen

3

24 g

of carbon and

1 g

of hydrogen

4

18 g

of carbon and

7 g

of hydrogen

Explanation:

Let the hydrocarbon is $C_{x} H_{y}$ .
$C_{x} H_{y} + O_{2} ⟶ {xCO}_{2} + \frac{y}{2} H_{2} O$
The weight of Carbon $= \frac{88}{44} \times 12$
$= 24 g$
The weight of hydrogen $= \frac{9}{18} \times 1$
$\begin{aligned} = 24 g \\ C_{x} H_{y} + (x + \frac{y}{4}) O_{2} \to {xCO}_{2} + \frac{y}{2} H_{2} O \\ C = x \times \frac{25}{M} = 2 \\ H = \frac{y}{2} \times \frac{25}{M} = 1 \\ \Rightarrow 24 : 1 ratio by mass \end{aligned}$

[JEE Main 2019

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