Question

In the laboratory, hydrogen gas is usually made by the following reaction: Zn(s) + 2 HCl(aq) → H2(g) + ZnCl2(aq) How many liters of H2 gas, collected over water at an atmospheric pressure of 752 mm Hg and a temperature of 21.0°C, can be made from 5.566 g of Zn and excess HCl? The partial pressure of water vapor is 18.65 mm Hg at 21.0°C.

Answer 1

Answer: The volume of hydrogen gas collected over water is 2.13 L

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of zinc = 5.566 g

Molar mass of zinc = 65.4 g/mol

Putting values in above equation, we get:

$\text{Moles of zinc}=\frac{5.566g}{65.4g/mol}=0.0851mol$

For the given chemical reaction:

$Zn(s)+2HCl(aq.)\rightarrow H_2(g)+ZnCl_2(aq.)$

As, HCl is present in excess. So, it is considered as an excess reagent.

Zinc is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of zinc produces 1 mole of hydrogen gas.

So, 0.0851 moles of zinc will produce = $\frac{1}[1}\times 0.0851=0.0851mol$ of hydrogen gas

To calculate the volume of hydrogen gas, we use ideal gas equation, which is:

PV = nRT

where,

P = Pressure of hydrogen gas = Total atmospheric pressure - vapor pressure of water = (752 - 18.65) mmHg = 733.35 mmHg

V = Volume of the hydrogen gas

n = number of moles of gas = 0.0851 moles

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

T = Temperature of hydrogen gas = $21^oC=[21+273]K=294K$

Putting values in above equation, we get:

$733.35mmHg\times V=0.0851mol\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 294K\\\\V=\frac{0.0851\times 62.3637\times 294}{733.35}=2.13L$

Hence, the volume of hydrogen gas collected over water is 2.13 L