Question

When active metals such as magnesium are immersed in acid solution, hydrogen gas is evolved. Calculate the volume(in L) of H2(g) at 35.2°C and 646 torr that can be formed when 325 mL of 0.855 M HCl solution reacts with excess Mg to give hydrogen gas and aqueous magnesium chloride.

Mg(s) + HCl(aq) --> MgCl2(aq) + H2(g) (unbalanced)

Answer 1

The volume of H₂ gas formed = 4.1 L

Explanation:

In the first step, we have to find the number of moles of HCl involved in the reaction

In the second step, using that moles of HCl, we have to find the moles of H₂.

In the third step, using the moles of H₂, we have to find the volume of H₂ by means of using the given temperature and pressure.

Number of moles of HCl = Molarity × volume

                            = 0.855 M × 0.325 L = 0.278 moles

We have to write the balanced equation as,

Mg(s) + 2 HCl(aq) → MgCl₂(aq) + H₂(g)

2 mol of HCl produces 1 mol of H₂

0.278 mol of HCl produces = $\frac{0.278 mol of HCl \times 1 mol of H_{2} }{2 mol HCl} = 0.139 mol of H_{2}$

Now we have to find volume of Hydrogen as,

PV = nRT

V = $\frac{nRT}{P}$

Plugin the given values as,

V = $\frac{0.139 mol \times 0.08205 \times 308.2}{0.85 atm} = 4.1 L$

So the volume of hydrogen formed is 4.1 L.