Question

Suppose we now collect hydrogen gas, H2(g), over water at 21◦C in a vessel with total pressure of 743 Torr. If the hydrogen gas is produced by the reaction of aluminum with hydrochloric acid:



2Al(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2(g)



what volume of hydrogen gas will be collected if 1.35 g Al(s) reacts with excess HCl(aq)? Express



your answer in liters.

Answer 1

Answer: The volume of hydrogen gas that will be collected is 1.85 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 aluminium = 1.35 g

Molar mass of aluminium = 27 g/mol

Putting values in above equation, we get:

$\text{Moles of aluminium}=\frac{1.35g}{27g/mol}=0.05mol$

For the given chemical reaction:

$2Al(s)+6HCl(aq.)\rightarrow 2AlCl_3(aq.)+3H_2(g)$

As, hydrochloric acid s present in excess. So, it is considered as an excess reagent.

Thus, aluminium is a limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

2 moles of aluminium produces 3 moles of hydrogen gas

So, 0.005 moles of aluminium will produce = $\frac{3}{2}\times 0.05=0.0750mol$ of hydrogen gas

To calculate the mass of helium gas, we use the equation given by ideal gas:

PV = nRT

where,

P = Pressure of hydrogen gas = 743 Torr

V = Volume of the helium gas = ?

n = number of moles of hydrogen gas = 0.075 mol

R = Gas constant = $62.364\text{ L Torr }mol^{-1}K^{-1}$

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

Putting values in above equation, we get:

$743Torr\times V=0.075mol\times 62.364\text{ L Torr }mol^{-1}K^{-1}\times 294K\\\\V=1.85L$

Hence, the volume of hydrogen gas that will be collected is 1.85 L

Answer 2

Answer:

$\boxed{\text{1.90 L}}$

Explanation:

1. Gather all the information in one place

M_r:   26.98

            2Al + 6HCl ⟶ 1AlCl₃ + 3H₂

m/g:     1.35

T = 21 °C

$\: \, p_{\text{tot}} = \text{743 torr}\\p_{\text{H2O}} = \text{18.7 torr}$

2. Moles of Al

$\text{Moles of Al} = \text{1.35 g Al} \times \dfrac{\text{1 mol Al}}{\text{26.98 g Al}} = \text{0.050 04 mol Al}$

3. Moles of H₂

The molar ratio is 3 mol H₂: 2 mol Al

$\text{Moles of H}_{2} = \text{0.050 04 mol Al} \times \dfrac{\text{3 mol H}_{2}}{\text{2 mol Al}} = \text{0.075 06 mol Al}$

4. Volume of H₂

We can use the Ideal Gas Law to calculate the volume of hydrogen.

The gas it collected over water, so the gas collected is a mixture of hydrogen and water vapour.

$p_{\text{H2}} = p_{\text{tot}} - p_{\text{H2O}} = 743 - 18.7 = \text{724.3 torr}\\\text{p}_{H2} = \text{724.3 torr} \times \dfrac{\text{1 atm}}{\text{760 torr}} = \text{0.9531 atm}$

T = 21 + 273.15 = 294.15 K

$\begin{array}{rcl}pV & = & nRT\\\text{0.95 31 atm} \times V & = & \text{0.075 06 mol} \times 0.082 06 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{294.15 K}\\0.95 31 V & = & \text{1.812 L}\\V & = & \textbf{1.90 L}\\\end{array}\\\text{The volume of hydrogen collected is \boxed{\textbf{1.90 L}} }$