QBManager

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307018 $X$ ' litres of carbon monoxide is present at STP. It is completely oxidized to ${CO}_{2}$ . The volume of ${CO}_{2}$ formed is 11.207 litres at STP. What is the value of ' $X$ ' in litres?

1 22.414

2 11.207

3 5.6035

4 44.828

Explanation:

$CO + \frac{1}{2} O_{2} \to {CO}_{2}$
1 mole of $CO$ gives 1 mole of ${CO}_{2}$
$∴ 11.207 L$ of CO at STP gives $11.207 L$ of ${CO}_{2}$ at STP.

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307019 Two litre of oxygen was reacted with five litre of CO at STP. The limiting reagent and mass of $C O_{2}$ produced are, respectively

1

C O, 1 .8 g

2

O_{2}, 2 .0 g

3

O_{2}, 7 .86 g

4

C O, 3 .93 g

Explanation:

$2 C O (g) + O_{2} (g) \to 2 C O_{2} (g)$
Initially, 5L, 2L
2 L of $O_{2}$ combines with 4 L of CO. Hence, $O_{2}$ is the limiting reagent.
When 2 L of $O_{2}$ combines with 4 L of CO, 4 L of $C O_{2}$ is formed.
22.4 L of $C O_{2}$ has the mass of 44 g
4 L of $C O_{2}$ has the mass of $\frac{4 \times 44}{22 .4} = 7 .86 g$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307020 How many moles of $P_{4} O_{6} a n d P_{4} O_{10}$ , respectively will be produced by the combustion of 12.4 g of phosphorous (atomic mass = 31) in 12.8 g of oxygen, leaving no $P_{4} o r O_{2}$ ?

1 0.1 mol and 0.3 mol

2 0.15 mol and 0.25 mol

3 0.05 mol each

4 0.1 mol each

Explanation:

$M o l e s o f P_{4} = \frac{12 .4}{4 \times 31} = 0 .1 m o l$
$M o l e s o f O_{2} = \frac{12 .8}{32} = 0 .4 m o l$
$L e t x m o l e s o f P_{4} f o r m P_{4} O_{6}$
$T h e n, m o l e s o f O_{2} r e q u i r e d$
$= x \times 3 + (0 .1 - x) \times 5 = 0 .4 (g i v e n) x = 0 .05$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307021 Calculate the weight of lime (CaO) obtained by heating 200 kg of 95% pure lime stone $(C a C O_{3})$ ?

1 104.4 kg

2 105.4 kg

3 212.8 kg

4 106.4 kg

Explanation:

100 kg impure sample has pure
$C a C O_{3} = 95 k g$
$200 k g$ impure sample has pure $C a C O_{3}$
$= \frac{95 \times 200}{100} = 190 k g$
$C a C O_{3} \to C a O + C O_{2}$
100 kg $C a C O_{3}$ gives 56 kg
$∴ 190 k g C a C O_{3} g i v e s$
$\frac{56 \times 190}{100} = 106 .4 k g o f C a O$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307018 $X$ ' litres of carbon monoxide is present at STP. It is completely oxidized to ${CO}_{2}$ . The volume of ${CO}_{2}$ formed is 11.207 litres at STP. What is the value of ' $X$ ' in litres?

1 22.414

2 11.207

3 5.6035

4 44.828

Explanation:

$CO + \frac{1}{2} O_{2} \to {CO}_{2}$
1 mole of $CO$ gives 1 mole of ${CO}_{2}$
$∴ 11.207 L$ of CO at STP gives $11.207 L$ of ${CO}_{2}$ at STP.

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307019 Two litre of oxygen was reacted with five litre of CO at STP. The limiting reagent and mass of $C O_{2}$ produced are, respectively

1

C O, 1 .8 g

2

O_{2}, 2 .0 g

3

O_{2}, 7 .86 g

4

C O, 3 .93 g

Explanation:

$2 C O (g) + O_{2} (g) \to 2 C O_{2} (g)$
Initially, 5L, 2L
2 L of $O_{2}$ combines with 4 L of CO. Hence, $O_{2}$ is the limiting reagent.
When 2 L of $O_{2}$ combines with 4 L of CO, 4 L of $C O_{2}$ is formed.
22.4 L of $C O_{2}$ has the mass of 44 g
4 L of $C O_{2}$ has the mass of $\frac{4 \times 44}{22 .4} = 7 .86 g$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307020 How many moles of $P_{4} O_{6} a n d P_{4} O_{10}$ , respectively will be produced by the combustion of 12.4 g of phosphorous (atomic mass = 31) in 12.8 g of oxygen, leaving no $P_{4} o r O_{2}$ ?

1 0.1 mol and 0.3 mol

2 0.15 mol and 0.25 mol

3 0.05 mol each

4 0.1 mol each

Explanation:

$M o l e s o f P_{4} = \frac{12 .4}{4 \times 31} = 0 .1 m o l$
$M o l e s o f O_{2} = \frac{12 .8}{32} = 0 .4 m o l$
$L e t x m o l e s o f P_{4} f o r m P_{4} O_{6}$
$T h e n, m o l e s o f O_{2} r e q u i r e d$
$= x \times 3 + (0 .1 - x) \times 5 = 0 .4 (g i v e n) x = 0 .05$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307021 Calculate the weight of lime (CaO) obtained by heating 200 kg of 95% pure lime stone $(C a C O_{3})$ ?

1 104.4 kg

2 105.4 kg

3 212.8 kg

4 106.4 kg

Explanation:

100 kg impure sample has pure
$C a C O_{3} = 95 k g$
$200 k g$ impure sample has pure $C a C O_{3}$
$= \frac{95 \times 200}{100} = 190 k g$
$C a C O_{3} \to C a O + C O_{2}$
100 kg $C a C O_{3}$ gives 56 kg
$∴ 190 k g C a C O_{3} g i v e s$
$\frac{56 \times 190}{100} = 106 .4 k g o f C a O$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307018 $X$ ' litres of carbon monoxide is present at STP. It is completely oxidized to ${CO}_{2}$ . The volume of ${CO}_{2}$ formed is 11.207 litres at STP. What is the value of ' $X$ ' in litres?

1 22.414

2 11.207

3 5.6035

4 44.828

Explanation:

$CO + \frac{1}{2} O_{2} \to {CO}_{2}$
1 mole of $CO$ gives 1 mole of ${CO}_{2}$
$∴ 11.207 L$ of CO at STP gives $11.207 L$ of ${CO}_{2}$ at STP.

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307019 Two litre of oxygen was reacted with five litre of CO at STP. The limiting reagent and mass of $C O_{2}$ produced are, respectively

1

C O, 1 .8 g

2

O_{2}, 2 .0 g

3

O_{2}, 7 .86 g

4

C O, 3 .93 g

Explanation:

$2 C O (g) + O_{2} (g) \to 2 C O_{2} (g)$
Initially, 5L, 2L
2 L of $O_{2}$ combines with 4 L of CO. Hence, $O_{2}$ is the limiting reagent.
When 2 L of $O_{2}$ combines with 4 L of CO, 4 L of $C O_{2}$ is formed.
22.4 L of $C O_{2}$ has the mass of 44 g
4 L of $C O_{2}$ has the mass of $\frac{4 \times 44}{22 .4} = 7 .86 g$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307020 How many moles of $P_{4} O_{6} a n d P_{4} O_{10}$ , respectively will be produced by the combustion of 12.4 g of phosphorous (atomic mass = 31) in 12.8 g of oxygen, leaving no $P_{4} o r O_{2}$ ?

1 0.1 mol and 0.3 mol

2 0.15 mol and 0.25 mol

3 0.05 mol each

4 0.1 mol each

Explanation:

$M o l e s o f P_{4} = \frac{12 .4}{4 \times 31} = 0 .1 m o l$
$M o l e s o f O_{2} = \frac{12 .8}{32} = 0 .4 m o l$
$L e t x m o l e s o f P_{4} f o r m P_{4} O_{6}$
$T h e n, m o l e s o f O_{2} r e q u i r e d$
$= x \times 3 + (0 .1 - x) \times 5 = 0 .4 (g i v e n) x = 0 .05$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307021 Calculate the weight of lime (CaO) obtained by heating 200 kg of 95% pure lime stone $(C a C O_{3})$ ?

1 104.4 kg

2 105.4 kg

3 212.8 kg

4 106.4 kg

Explanation:

100 kg impure sample has pure
$C a C O_{3} = 95 k g$
$200 k g$ impure sample has pure $C a C O_{3}$
$= \frac{95 \times 200}{100} = 190 k g$
$C a C O_{3} \to C a O + C O_{2}$
100 kg $C a C O_{3}$ gives 56 kg
$∴ 190 k g C a C O_{3} g i v e s$
$\frac{56 \times 190}{100} = 106 .4 k g o f C a O$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307018 $X$ ' litres of carbon monoxide is present at STP. It is completely oxidized to ${CO}_{2}$ . The volume of ${CO}_{2}$ formed is 11.207 litres at STP. What is the value of ' $X$ ' in litres?

1 22.414

2 11.207

3 5.6035

4 44.828

Explanation:

$CO + \frac{1}{2} O_{2} \to {CO}_{2}$
1 mole of $CO$ gives 1 mole of ${CO}_{2}$
$∴ 11.207 L$ of CO at STP gives $11.207 L$ of ${CO}_{2}$ at STP.

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307019 Two litre of oxygen was reacted with five litre of CO at STP. The limiting reagent and mass of $C O_{2}$ produced are, respectively

1

C O, 1 .8 g

2

O_{2}, 2 .0 g

3

O_{2}, 7 .86 g

4

C O, 3 .93 g

Explanation:

$2 C O (g) + O_{2} (g) \to 2 C O_{2} (g)$
Initially, 5L, 2L
2 L of $O_{2}$ combines with 4 L of CO. Hence, $O_{2}$ is the limiting reagent.
When 2 L of $O_{2}$ combines with 4 L of CO, 4 L of $C O_{2}$ is formed.
22.4 L of $C O_{2}$ has the mass of 44 g
4 L of $C O_{2}$ has the mass of $\frac{4 \times 44}{22 .4} = 7 .86 g$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307020 How many moles of $P_{4} O_{6} a n d P_{4} O_{10}$ , respectively will be produced by the combustion of 12.4 g of phosphorous (atomic mass = 31) in 12.8 g of oxygen, leaving no $P_{4} o r O_{2}$ ?

1 0.1 mol and 0.3 mol

2 0.15 mol and 0.25 mol

3 0.05 mol each

4 0.1 mol each

Explanation:

$M o l e s o f P_{4} = \frac{12 .4}{4 \times 31} = 0 .1 m o l$
$M o l e s o f O_{2} = \frac{12 .8}{32} = 0 .4 m o l$
$L e t x m o l e s o f P_{4} f o r m P_{4} O_{6}$
$T h e n, m o l e s o f O_{2} r e q u i r e d$
$= x \times 3 + (0 .1 - x) \times 5 = 0 .4 (g i v e n) x = 0 .05$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307021 Calculate the weight of lime (CaO) obtained by heating 200 kg of 95% pure lime stone $(C a C O_{3})$ ?

1 104.4 kg

2 105.4 kg

3 212.8 kg

4 106.4 kg

Explanation:

100 kg impure sample has pure
$C a C O_{3} = 95 k g$
$200 k g$ impure sample has pure $C a C O_{3}$
$= \frac{95 \times 200}{100} = 190 k g$
$C a C O_{3} \to C a O + C O_{2}$
100 kg $C a C O_{3}$ gives 56 kg
$∴ 190 k g C a C O_{3} g i v e s$
$\frac{56 \times 190}{100} = 106 .4 k g o f C a O$

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: