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CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307026 1 gram of a carbonate $(M_{2} C O_{3})$ on treatment with excess HCl produces 0.01186 mole of $C O_{2}$ . The molar mass of $M_{2} C O_{3} i n g m o l^{- 1}$ is:

1 1186

2 84.3

3 118.6

4 11.86

Explanation:

Given chemical equation,
$\underset{1 g m}{M_{2}} C O_{3} + 2 H C l \to 2 M C l + H_{2} O + \underset{0 .01186 m o l}{C O_{2}}$
from the balanced chemical equation,
$n_{M_{2} C O_{3}} = n_{C O_{2}}$
$\frac{1}{M o l a r m a s s o f M_{2} C O_{3}} = 0 .01186$
$M o l a r m a s s o f M_{2} C O_{3} = \frac{1}{0 .01186} = 53 .76$
$∴ 2 M + 12 + 48 = 53 .7$
$M = 84 .3 g m / m o l$ .

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307027 What is the weight of oxygen required for the complete combustion of 2.8 kg of ethylene?

1 2.8 kg

2 6.4 kg

3 9.6 kg

4 96 kg

Explanation:

$C_{2} H_{4} + 3 O_{2} \to 2 C O_{2} + 2 H_{2} O$
28 kg of $C_{2} H_{4}$ gives 96 kg of $O_{2}$
2.8 kg of $C_{2} H_{4}$ gives $\frac{96 \times 2 .8}{28} k g$ of $O_{2}$
$= 9 .6 k g o f O_{2}$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307028 When 1.25 gm of a chalk sample $(C a C O_{3})$ is strongly heated, 0.44 gm $C O_{2}$ gas is produced. Determine the percentage yield of the reaction.

1

40 %

2

60 %

3

80 %

4

90 %

Explanation:

$\underset{\underset{1 .25 g}{100 g}}{CaC O_{3}} \to CaO + \underset{\underset{?}{44 g}}{C O_{2}}$
Amount of $C O_{2}$ formed $= \frac{1 .25 \times 44}{100} = 0 .55 g$
$B u t 0 .44 g o f C O_{2} f o r m e d$
$S o, % y i e l d = \frac{0 .44 \times 100}{0 .55} = 80 %$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307029 Match Column I with Column II and choose the correct combination from the given options given.
supporting img

1

A - R, B - P, C - S, D - Q

2

A - R, B - S, C - P, D - Q

3

A - P, B - R, C - Q, D - S

4

A - S, B - P, C - R, D - Q

Explanation:

$(1) \underset{1.0 g}{2 H_{2}} + \underset{1 .0 g}{O_{2}} \to 2 H_{2} O$
$\frac{1}{2} m o l \frac{1}{32} m o l$
$O_{2}$ is the limiting reagent :
$1 mol O_{2} = 2 mol H_{2} O$
$\Rightarrow \frac{1}{32} mol O_{2} \equiv \frac{1}{16} mol H_{2} O$
$= \frac{18}{16} g H_{2} O = 1.125 g H_{2} O$
$(2) \underset{1 .0 g}{N_{2}} + \underset{1 .0 g}{3 H_{2}} \to 2 N H_{3}$
$\frac{1}{28} m o l \frac{1}{2} m o l$
$N_{2}$ is the limiting reagent.
$∴ \frac{1}{28} m o l N_{2} \equiv \frac{1}{14} m o l$
$N H_{3} \equiv \frac{1}{14} \times 17 g = 1.214 g$
$(3) \underset{1 .0 g}{CaC O_{3}} \overset{Δ}{\to} \underset{?}{CaO} + C O_{2}$
$\frac{1}{100} mol CaC O_{3}$
$\frac{1}{100} \times 56 g CaO = 0.56 g CaO$
$(4) \underset{1.0 g}{C} + \underset{1.0 g}{2 H_{2}} \to C H_{4}$
$\frac{1}{12} m o l \frac{1}{2} m o l$
C is the limiting reagent.
$∴ \frac{1}{12} mol C \equiv \frac{1}{12} mol$
$C H_{4} \equiv \frac{16}{12} g = 1.333 g C H_{4}$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307030 A metal oxide has the formula $Z_{2} O_{3}$ . It can be reduced by hydrogen to give free metal and water. 0.1596 g of the metal oxide requires 6 mg of hydrogen for complete reduction. The atomic weight of the metal is

1 27.9

2 159.6

3 79.8

4 55.8

Explanation:

Valency of metal in $Z_{2} O_{3} = 3$
$Z_{2} O_{3} + 3 H_{2} \to 2 Z + 3 H_{2} O$
0.1596 g of $Z_{2} O_{3}$ react with $H_{2} = 6 m g = 0 .006 g$
$∴$ $1 g o f H_{2}$ react with $Z_{2} O_{3} = \frac{0 .1596}{0 .006} = 26 .6 g$
$∴$ Eq. wt. of $Z_{2} O_{3} = 26 .6$
Eq. wt. of $Z + E q w t . o f O = E + 8$
$E + 8 = 26 .6$ or $E = 18 .6$
$∴ A t . w t . o f Z = 18 .6 \times 3 = 55 .8$ .

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307026 1 gram of a carbonate $(M_{2} C O_{3})$ on treatment with excess HCl produces 0.01186 mole of $C O_{2}$ . The molar mass of $M_{2} C O_{3} i n g m o l^{- 1}$ is:

1 1186

2 84.3

3 118.6

4 11.86

Explanation:

Given chemical equation,
$\underset{1 g m}{M_{2}} C O_{3} + 2 H C l \to 2 M C l + H_{2} O + \underset{0 .01186 m o l}{C O_{2}}$
from the balanced chemical equation,
$n_{M_{2} C O_{3}} = n_{C O_{2}}$
$\frac{1}{M o l a r m a s s o f M_{2} C O_{3}} = 0 .01186$
$M o l a r m a s s o f M_{2} C O_{3} = \frac{1}{0 .01186} = 53 .76$
$∴ 2 M + 12 + 48 = 53 .7$
$M = 84 .3 g m / m o l$ .

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307027 What is the weight of oxygen required for the complete combustion of 2.8 kg of ethylene?

1 2.8 kg

2 6.4 kg

3 9.6 kg

4 96 kg

Explanation:

$C_{2} H_{4} + 3 O_{2} \to 2 C O_{2} + 2 H_{2} O$
28 kg of $C_{2} H_{4}$ gives 96 kg of $O_{2}$
2.8 kg of $C_{2} H_{4}$ gives $\frac{96 \times 2 .8}{28} k g$ of $O_{2}$
$= 9 .6 k g o f O_{2}$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307028 When 1.25 gm of a chalk sample $(C a C O_{3})$ is strongly heated, 0.44 gm $C O_{2}$ gas is produced. Determine the percentage yield of the reaction.

1

40 %

2

60 %

3

80 %

4

90 %

Explanation:

$\underset{\underset{1 .25 g}{100 g}}{CaC O_{3}} \to CaO + \underset{\underset{?}{44 g}}{C O_{2}}$
Amount of $C O_{2}$ formed $= \frac{1 .25 \times 44}{100} = 0 .55 g$
$B u t 0 .44 g o f C O_{2} f o r m e d$
$S o, % y i e l d = \frac{0 .44 \times 100}{0 .55} = 80 %$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307029 Match Column I with Column II and choose the correct combination from the given options given.
supporting img

1

A - R, B - P, C - S, D - Q

2

A - R, B - S, C - P, D - Q

3

A - P, B - R, C - Q, D - S

4

A - S, B - P, C - R, D - Q

Explanation:

$(1) \underset{1.0 g}{2 H_{2}} + \underset{1 .0 g}{O_{2}} \to 2 H_{2} O$
$\frac{1}{2} m o l \frac{1}{32} m o l$
$O_{2}$ is the limiting reagent :
$1 mol O_{2} = 2 mol H_{2} O$
$\Rightarrow \frac{1}{32} mol O_{2} \equiv \frac{1}{16} mol H_{2} O$
$= \frac{18}{16} g H_{2} O = 1.125 g H_{2} O$
$(2) \underset{1 .0 g}{N_{2}} + \underset{1 .0 g}{3 H_{2}} \to 2 N H_{3}$
$\frac{1}{28} m o l \frac{1}{2} m o l$
$N_{2}$ is the limiting reagent.
$∴ \frac{1}{28} m o l N_{2} \equiv \frac{1}{14} m o l$
$N H_{3} \equiv \frac{1}{14} \times 17 g = 1.214 g$
$(3) \underset{1 .0 g}{CaC O_{3}} \overset{Δ}{\to} \underset{?}{CaO} + C O_{2}$
$\frac{1}{100} mol CaC O_{3}$
$\frac{1}{100} \times 56 g CaO = 0.56 g CaO$
$(4) \underset{1.0 g}{C} + \underset{1.0 g}{2 H_{2}} \to C H_{4}$
$\frac{1}{12} m o l \frac{1}{2} m o l$
C is the limiting reagent.
$∴ \frac{1}{12} mol C \equiv \frac{1}{12} mol$
$C H_{4} \equiv \frac{16}{12} g = 1.333 g C H_{4}$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307030 A metal oxide has the formula $Z_{2} O_{3}$ . It can be reduced by hydrogen to give free metal and water. 0.1596 g of the metal oxide requires 6 mg of hydrogen for complete reduction. The atomic weight of the metal is

1 27.9

2 159.6

3 79.8

4 55.8

Explanation:

Valency of metal in $Z_{2} O_{3} = 3$
$Z_{2} O_{3} + 3 H_{2} \to 2 Z + 3 H_{2} O$
0.1596 g of $Z_{2} O_{3}$ react with $H_{2} = 6 m g = 0 .006 g$
$∴$ $1 g o f H_{2}$ react with $Z_{2} O_{3} = \frac{0 .1596}{0 .006} = 26 .6 g$
$∴$ Eq. wt. of $Z_{2} O_{3} = 26 .6$
Eq. wt. of $Z + E q w t . o f O = E + 8$
$E + 8 = 26 .6$ or $E = 18 .6$
$∴ A t . w t . o f Z = 18 .6 \times 3 = 55 .8$ .

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307026 1 gram of a carbonate $(M_{2} C O_{3})$ on treatment with excess HCl produces 0.01186 mole of $C O_{2}$ . The molar mass of $M_{2} C O_{3} i n g m o l^{- 1}$ is:

1 1186

2 84.3

3 118.6

4 11.86

Explanation:

Given chemical equation,
$\underset{1 g m}{M_{2}} C O_{3} + 2 H C l \to 2 M C l + H_{2} O + \underset{0 .01186 m o l}{C O_{2}}$
from the balanced chemical equation,
$n_{M_{2} C O_{3}} = n_{C O_{2}}$
$\frac{1}{M o l a r m a s s o f M_{2} C O_{3}} = 0 .01186$
$M o l a r m a s s o f M_{2} C O_{3} = \frac{1}{0 .01186} = 53 .76$
$∴ 2 M + 12 + 48 = 53 .7$
$M = 84 .3 g m / m o l$ .

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307027 What is the weight of oxygen required for the complete combustion of 2.8 kg of ethylene?

1 2.8 kg

2 6.4 kg

3 9.6 kg

4 96 kg

Explanation:

$C_{2} H_{4} + 3 O_{2} \to 2 C O_{2} + 2 H_{2} O$
28 kg of $C_{2} H_{4}$ gives 96 kg of $O_{2}$
2.8 kg of $C_{2} H_{4}$ gives $\frac{96 \times 2 .8}{28} k g$ of $O_{2}$
$= 9 .6 k g o f O_{2}$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307028 When 1.25 gm of a chalk sample $(C a C O_{3})$ is strongly heated, 0.44 gm $C O_{2}$ gas is produced. Determine the percentage yield of the reaction.

1

40 %

2

60 %

3

80 %

4

90 %

Explanation:

$\underset{\underset{1 .25 g}{100 g}}{CaC O_{3}} \to CaO + \underset{\underset{?}{44 g}}{C O_{2}}$
Amount of $C O_{2}$ formed $= \frac{1 .25 \times 44}{100} = 0 .55 g$
$B u t 0 .44 g o f C O_{2} f o r m e d$
$S o, % y i e l d = \frac{0 .44 \times 100}{0 .55} = 80 %$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307029 Match Column I with Column II and choose the correct combination from the given options given.
supporting img

1

A - R, B - P, C - S, D - Q

2

A - R, B - S, C - P, D - Q

3

A - P, B - R, C - Q, D - S

4

A - S, B - P, C - R, D - Q

Explanation:

$(1) \underset{1.0 g}{2 H_{2}} + \underset{1 .0 g}{O_{2}} \to 2 H_{2} O$
$\frac{1}{2} m o l \frac{1}{32} m o l$
$O_{2}$ is the limiting reagent :
$1 mol O_{2} = 2 mol H_{2} O$
$\Rightarrow \frac{1}{32} mol O_{2} \equiv \frac{1}{16} mol H_{2} O$
$= \frac{18}{16} g H_{2} O = 1.125 g H_{2} O$
$(2) \underset{1 .0 g}{N_{2}} + \underset{1 .0 g}{3 H_{2}} \to 2 N H_{3}$
$\frac{1}{28} m o l \frac{1}{2} m o l$
$N_{2}$ is the limiting reagent.
$∴ \frac{1}{28} m o l N_{2} \equiv \frac{1}{14} m o l$
$N H_{3} \equiv \frac{1}{14} \times 17 g = 1.214 g$
$(3) \underset{1 .0 g}{CaC O_{3}} \overset{Δ}{\to} \underset{?}{CaO} + C O_{2}$
$\frac{1}{100} mol CaC O_{3}$
$\frac{1}{100} \times 56 g CaO = 0.56 g CaO$
$(4) \underset{1.0 g}{C} + \underset{1.0 g}{2 H_{2}} \to C H_{4}$
$\frac{1}{12} m o l \frac{1}{2} m o l$
C is the limiting reagent.
$∴ \frac{1}{12} mol C \equiv \frac{1}{12} mol$
$C H_{4} \equiv \frac{16}{12} g = 1.333 g C H_{4}$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307030 A metal oxide has the formula $Z_{2} O_{3}$ . It can be reduced by hydrogen to give free metal and water. 0.1596 g of the metal oxide requires 6 mg of hydrogen for complete reduction. The atomic weight of the metal is

1 27.9

2 159.6

3 79.8

4 55.8

Explanation:

Valency of metal in $Z_{2} O_{3} = 3$
$Z_{2} O_{3} + 3 H_{2} \to 2 Z + 3 H_{2} O$
0.1596 g of $Z_{2} O_{3}$ react with $H_{2} = 6 m g = 0 .006 g$
$∴$ $1 g o f H_{2}$ react with $Z_{2} O_{3} = \frac{0 .1596}{0 .006} = 26 .6 g$
$∴$ Eq. wt. of $Z_{2} O_{3} = 26 .6$
Eq. wt. of $Z + E q w t . o f O = E + 8$
$E + 8 = 26 .6$ or $E = 18 .6$
$∴ A t . w t . o f Z = 18 .6 \times 3 = 55 .8$ .

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307026 1 gram of a carbonate $(M_{2} C O_{3})$ on treatment with excess HCl produces 0.01186 mole of $C O_{2}$ . The molar mass of $M_{2} C O_{3} i n g m o l^{- 1}$ is:

1 1186

2 84.3

3 118.6

4 11.86

Explanation:

Given chemical equation,
$\underset{1 g m}{M_{2}} C O_{3} + 2 H C l \to 2 M C l + H_{2} O + \underset{0 .01186 m o l}{C O_{2}}$
from the balanced chemical equation,
$n_{M_{2} C O_{3}} = n_{C O_{2}}$
$\frac{1}{M o l a r m a s s o f M_{2} C O_{3}} = 0 .01186$
$M o l a r m a s s o f M_{2} C O_{3} = \frac{1}{0 .01186} = 53 .76$
$∴ 2 M + 12 + 48 = 53 .7$
$M = 84 .3 g m / m o l$ .

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307027 What is the weight of oxygen required for the complete combustion of 2.8 kg of ethylene?

1 2.8 kg

2 6.4 kg

3 9.6 kg

4 96 kg

Explanation:

$C_{2} H_{4} + 3 O_{2} \to 2 C O_{2} + 2 H_{2} O$
28 kg of $C_{2} H_{4}$ gives 96 kg of $O_{2}$
2.8 kg of $C_{2} H_{4}$ gives $\frac{96 \times 2 .8}{28} k g$ of $O_{2}$
$= 9 .6 k g o f O_{2}$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307028 When 1.25 gm of a chalk sample $(C a C O_{3})$ is strongly heated, 0.44 gm $C O_{2}$ gas is produced. Determine the percentage yield of the reaction.

1

40 %

2

60 %

3

80 %

4

90 %

Explanation:

$\underset{\underset{1 .25 g}{100 g}}{CaC O_{3}} \to CaO + \underset{\underset{?}{44 g}}{C O_{2}}$
Amount of $C O_{2}$ formed $= \frac{1 .25 \times 44}{100} = 0 .55 g$
$B u t 0 .44 g o f C O_{2} f o r m e d$
$S o, % y i e l d = \frac{0 .44 \times 100}{0 .55} = 80 %$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307029 Match Column I with Column II and choose the correct combination from the given options given.
supporting img

1

A - R, B - P, C - S, D - Q

2

A - R, B - S, C - P, D - Q

3

A - P, B - R, C - Q, D - S

4

A - S, B - P, C - R, D - Q

Explanation:

$(1) \underset{1.0 g}{2 H_{2}} + \underset{1 .0 g}{O_{2}} \to 2 H_{2} O$
$\frac{1}{2} m o l \frac{1}{32} m o l$
$O_{2}$ is the limiting reagent :
$1 mol O_{2} = 2 mol H_{2} O$
$\Rightarrow \frac{1}{32} mol O_{2} \equiv \frac{1}{16} mol H_{2} O$
$= \frac{18}{16} g H_{2} O = 1.125 g H_{2} O$
$(2) \underset{1 .0 g}{N_{2}} + \underset{1 .0 g}{3 H_{2}} \to 2 N H_{3}$
$\frac{1}{28} m o l \frac{1}{2} m o l$
$N_{2}$ is the limiting reagent.
$∴ \frac{1}{28} m o l N_{2} \equiv \frac{1}{14} m o l$
$N H_{3} \equiv \frac{1}{14} \times 17 g = 1.214 g$
$(3) \underset{1 .0 g}{CaC O_{3}} \overset{Δ}{\to} \underset{?}{CaO} + C O_{2}$
$\frac{1}{100} mol CaC O_{3}$
$\frac{1}{100} \times 56 g CaO = 0.56 g CaO$
$(4) \underset{1.0 g}{C} + \underset{1.0 g}{2 H_{2}} \to C H_{4}$
$\frac{1}{12} m o l \frac{1}{2} m o l$
C is the limiting reagent.
$∴ \frac{1}{12} mol C \equiv \frac{1}{12} mol$
$C H_{4} \equiv \frac{16}{12} g = 1.333 g C H_{4}$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307030 A metal oxide has the formula $Z_{2} O_{3}$ . It can be reduced by hydrogen to give free metal and water. 0.1596 g of the metal oxide requires 6 mg of hydrogen for complete reduction. The atomic weight of the metal is

1 27.9

2 159.6

3 79.8

4 55.8

Explanation:

Valency of metal in $Z_{2} O_{3} = 3$
$Z_{2} O_{3} + 3 H_{2} \to 2 Z + 3 H_{2} O$
0.1596 g of $Z_{2} O_{3}$ react with $H_{2} = 6 m g = 0 .006 g$
$∴$ $1 g o f H_{2}$ react with $Z_{2} O_{3} = \frac{0 .1596}{0 .006} = 26 .6 g$
$∴$ Eq. wt. of $Z_{2} O_{3} = 26 .6$
Eq. wt. of $Z + E q w t . o f O = E + 8$
$E + 8 = 26 .6$ or $E = 18 .6$
$∴ A t . w t . o f Z = 18 .6 \times 3 = 55 .8$ .

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307026 1 gram of a carbonate $(M_{2} C O_{3})$ on treatment with excess HCl produces 0.01186 mole of $C O_{2}$ . The molar mass of $M_{2} C O_{3} i n g m o l^{- 1}$ is:

1 1186

2 84.3

3 118.6

4 11.86

Explanation:

Given chemical equation,
$\underset{1 g m}{M_{2}} C O_{3} + 2 H C l \to 2 M C l + H_{2} O + \underset{0 .01186 m o l}{C O_{2}}$
from the balanced chemical equation,
$n_{M_{2} C O_{3}} = n_{C O_{2}}$
$\frac{1}{M o l a r m a s s o f M_{2} C O_{3}} = 0 .01186$
$M o l a r m a s s o f M_{2} C O_{3} = \frac{1}{0 .01186} = 53 .76$
$∴ 2 M + 12 + 48 = 53 .7$
$M = 84 .3 g m / m o l$ .

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307027 What is the weight of oxygen required for the complete combustion of 2.8 kg of ethylene?

1 2.8 kg

2 6.4 kg

3 9.6 kg

4 96 kg

Explanation:

$C_{2} H_{4} + 3 O_{2} \to 2 C O_{2} + 2 H_{2} O$
28 kg of $C_{2} H_{4}$ gives 96 kg of $O_{2}$
2.8 kg of $C_{2} H_{4}$ gives $\frac{96 \times 2 .8}{28} k g$ of $O_{2}$
$= 9 .6 k g o f O_{2}$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307028 When 1.25 gm of a chalk sample $(C a C O_{3})$ is strongly heated, 0.44 gm $C O_{2}$ gas is produced. Determine the percentage yield of the reaction.

1

40 %

2

60 %

3

80 %

4

90 %

Explanation:

$\underset{\underset{1 .25 g}{100 g}}{CaC O_{3}} \to CaO + \underset{\underset{?}{44 g}}{C O_{2}}$
Amount of $C O_{2}$ formed $= \frac{1 .25 \times 44}{100} = 0 .55 g$
$B u t 0 .44 g o f C O_{2} f o r m e d$
$S o, % y i e l d = \frac{0 .44 \times 100}{0 .55} = 80 %$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307029 Match Column I with Column II and choose the correct combination from the given options given.
supporting img

1

A - R, B - P, C - S, D - Q

2

A - R, B - S, C - P, D - Q

3

A - P, B - R, C - Q, D - S

4

A - S, B - P, C - R, D - Q

Explanation:

$(1) \underset{1.0 g}{2 H_{2}} + \underset{1 .0 g}{O_{2}} \to 2 H_{2} O$
$\frac{1}{2} m o l \frac{1}{32} m o l$
$O_{2}$ is the limiting reagent :
$1 mol O_{2} = 2 mol H_{2} O$
$\Rightarrow \frac{1}{32} mol O_{2} \equiv \frac{1}{16} mol H_{2} O$
$= \frac{18}{16} g H_{2} O = 1.125 g H_{2} O$
$(2) \underset{1 .0 g}{N_{2}} + \underset{1 .0 g}{3 H_{2}} \to 2 N H_{3}$
$\frac{1}{28} m o l \frac{1}{2} m o l$
$N_{2}$ is the limiting reagent.
$∴ \frac{1}{28} m o l N_{2} \equiv \frac{1}{14} m o l$
$N H_{3} \equiv \frac{1}{14} \times 17 g = 1.214 g$
$(3) \underset{1 .0 g}{CaC O_{3}} \overset{Δ}{\to} \underset{?}{CaO} + C O_{2}$
$\frac{1}{100} mol CaC O_{3}$
$\frac{1}{100} \times 56 g CaO = 0.56 g CaO$
$(4) \underset{1.0 g}{C} + \underset{1.0 g}{2 H_{2}} \to C H_{4}$
$\frac{1}{12} m o l \frac{1}{2} m o l$
C is the limiting reagent.
$∴ \frac{1}{12} mol C \equiv \frac{1}{12} mol$
$C H_{4} \equiv \frac{16}{12} g = 1.333 g C H_{4}$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307030 A metal oxide has the formula $Z_{2} O_{3}$ . It can be reduced by hydrogen to give free metal and water. 0.1596 g of the metal oxide requires 6 mg of hydrogen for complete reduction. The atomic weight of the metal is

1 27.9

2 159.6

3 79.8

4 55.8

Explanation:

Valency of metal in $Z_{2} O_{3} = 3$
$Z_{2} O_{3} + 3 H_{2} \to 2 Z + 3 H_{2} O$
0.1596 g of $Z_{2} O_{3}$ react with $H_{2} = 6 m g = 0 .006 g$
$∴$ $1 g o f H_{2}$ react with $Z_{2} O_{3} = \frac{0 .1596}{0 .006} = 26 .6 g$
$∴$ Eq. wt. of $Z_{2} O_{3} = 26 .6$
Eq. wt. of $Z + E q w t . o f O = E + 8$
$E + 8 = 26 .6$ or $E = 18 .6$
$∴ A t . w t . o f Z = 18 .6 \times 3 = 55 .8$ .