Question

Consider the following reaction :
Zn(s) + 2HCl (aq) → ZnCl2(aq) + H2(9)
If 520 mL of H (9) is collected over water at 28 °C
and the atmospheric pressure is 1 atm. If vapour
pressure of water at 28°C is 28.3 mm Hg. Then the
amount of Zn (in g) taken at the start of the
reaction is (assuming all the Zn(s) gets converted
in product, molar mass of Zn is 65 g/mol]​

Answer 1

Explanation:

1 mole Zn→ 1 mole H

2

218

65.4gm

Zn→

218

1mole

H

2

0.3 gm Zn→

48

1

mole H

2

⇒ 1 mole H

2

→22.4L

218

1

mole H

2

→

218

22.4

L=0.10275 litres

Answer 2

Answer:

The amount of zinc in grams needed at the start of the reaction, m = $1.3g$.

Explanation:

Given data,

The reaction: $Zn(s) + 2HCl(aq) \rightarrow ZnCl_{2} + H_{2}$

The volume of hydrogen gas over water, $V_{H_{2} }$ = $520ml$ = $0.52L$

The temperature of hydrogen gas, $T_{H_{2} }$ = $28\textdegree C$ = $(28+273.15)K$ = $301.15K$

The given atmospheric pressure, P = $1atm$

The pressure of water, $P_{water}$ = $28.3mmHg$

Convert mmHg in atm

• $1 mmHg$ = $0.00131atm$
• $28.3 mmHg$ = $(1 mmHg\times 0.00131)atm$ = $0.037atm$

The molar mass of zinc = $65g/mol$

The amount of zinc in grams needed at the start of the reaction, m =?

Firstly, we have to calculate the pressure of the hydrogen gas:

• $P_{H_{2}gas }$ = $P - P_{water}$ = $1 - 0.037$ = $0.963atm$

Now, we have to calculate the number of moles of hydrogen gas formed in the reaction by using the ideal gas equation:

• $PV =nRT$
• $n = \frac{PV}{RT}$

Here, R = gas constant = $0.082L-atm/K-mol$

For hydrogen gas;

• n = $\frac{0.963\times 0.52}{0.082\times 301.15}$
• n = $\frac{0.5007}{24.69}$ = $0.02mol$

Now, as given in the reaction:

• $Zn(s) + 2HCl(aq) \rightarrow ZnCl_{2} + H_{2}$
• 1 mole of Zinc formed 1 mole of Hydrogen gas.

Therefore,

• Number of moles of zinc = number of moles of hydrogen gas
• Number of moles of zinc = $0.02mol$

As we know,

• Number of moles = $\frac{mass}{molar mass}$

For zinc,

• Mass = $0.02\times 65$ = $1.3g$

Hence, the amount of zinc in grams needed at the start of the reaction, m = $1.3g$.