## Jan 28, 2017

Q-7)

The balanced equation of the given reaction

$F {e}_{2} {O}_{3} + 3 C O \to 2 F e + 3 C {O}_{2}$

Molar mass of $F {e}_{2} {O}_{3} = \left(2 \times 56 + 3 \times 16\right) = 160 g \text{/mol}$

Reacting mass of $F {e}_{2} {O}_{3} = 1 k g = 1000 g = \frac{1000 g}{160 g \text{/mol}} = 6.25 m o l$

Molar mass of $C O = 12 + 16 = 28 g \text{/mol}$

Let reacting mass of $C O = x g = \frac{x g}{28 g \text{/mol}} = \frac{x}{28} m o l$

As per balanced equation the reacting mole ratio of

$F {e}_{2} {O}_{3} : C O = 1 : 3$

So $\frac{6.25}{\frac{x}{28}} = \frac{1}{3}$

$\implies x = 3 \times 6.25 \times 28 g = 525 g = 0.625 k g$

Q-8)

The balanced equation of the given reaction

$2 F e + 3 C {l}_{2} \to 2 F e C {l}_{3}$

Molar mass of $F e = 56 g \text{/mol}$

Reacting mass of $F e = 20 g = \frac{20 g}{56 g \text{/mol}} = \frac{5}{14} m o l$

Molar mass of $C {l}_{2} = 2 \times 35.5 = 71 g \text{/mol}$

Let reacting mass of $C O = y g = \frac{y g}{71 g \text{/mol}} = \frac{y}{71} m o l$

As per balanced equation the reacting mole ratio of

$F e : C {l}_{2} = 2 : 3$

So $\frac{\frac{5}{14}}{\frac{y}{71}} = \frac{2}{3}$

$\implies y = \frac{3 \times 5 \times 71}{2 \times 14} g = 38.03 g$

Q-9)

The balanced equation of the given reaction

$2 N {H}_{3} + {H}_{2} {O}_{2} \to {N}_{2} {H}_{4} + {H}_{2} O$

Molar mass of $N {H}_{3} = 14 + 3 \times 1 = 17 g \text{/mol}$

Let the required reacting mass of

$N {H}_{3} = z g = \frac{z g}{17 g \text{/mol}} = \frac{z}{17} m o l$

Molar mass of ${N}_{2} {H}_{4} = 2 \times 14 + 4 \times 1 = 32 g \text{/mol}$

The produced mass of ${N}_{2} {H}_{4} = 100 g = \frac{100 g}{32 g \text{/mol}} = \frac{25}{8} m o l$

As per balanced equation the mole ratio of

$N {H}_{3} : {N}_{2} {H}_{4} = 2 : 1$

So $\frac{\frac{z}{17}}{\frac{25}{8}} = \frac{2}{1}$

$\implies z = \frac{2 \times 25 \times 17}{8} g = 106.25 g$

Q-10)

The balanced equation of the physical change for the dehydration of hydrated $N {a}_{2} S {O}_{4.} x {H}_{2} O$ is given as follows

$N {a}_{2} S {O}_{4.} x {H}_{2} O \to N {a}_{2} S {O}_{4} + x {H}_{2} O$

Formula mass of $N {a}_{2} S {O}_{4.} x {H}_{2} O$

$= \left(2 \times 23 + 32 + 4 \times 16 + x \left(2 \times 1 + 16\right)\right) g \text{/mol}$

$= \left(142 + 18 x\right) g \text{/mol}$

Formula mass of anhydrous $N {a}_{2} S {O}_{4}$

$= \left(2 \times 23 + 32 + 4 \times 16\right) g \text{/mol}$

$= 142 g \text{/mol}$

It is obvious from the equation that

$\frac{142 + 18 x}{142} = \frac{10}{4.4} = \frac{25}{11}$

$\implies 18 x = \frac{25 \times 142}{11} - 142 = \frac{14 \times 142}{11}$

$\implies x = \frac{14 \times 142}{11 \times 18} \approx 10$

So Formula mass of $N {a}_{2} S {O}_{4.} 10 {H}_{2} O$

$= \left(2 \times 23 + 32 + 4 \times 16 + 10 \times \left(2 \times 1 + 16\right)\right) g \text{/mol}$

$= 322 g \text{/mol}$