Chapter 25, Problem 105A

Interpretation Introduction

Interpretation:

The piece of magnesium with a mass of $\text{10}\text{.00g}$ is added to sulphuric acid. The no. of moles and volume of hydrogen gas generated is to be determined.

Concept introduction:

An acid reacts with metals to produce hydrogen gas and metal salts. Magnesium is a metal that reacts quickly. Magnesium chloride and hydrogen gas will be produced during the reaction with hydrochloric acid.

Answer to Problem 105A

Moles of hydrogen gas = $\text{0}{\text{.412 mol H}}_{\text{2}}$

The hydrogen gas produced if the magnesium reacts completely = $\text{9}{\text{.23×10}}^{\text{3}}{\text{cm}}^{\text{3}}{\text{H}}_{\text{2}}$

Explanation of Solution

In order to solve the question:

First, establish the balanced chemical reaction between magnesium and sulphuric acid.

  ${\text{Mg+H}}_{\text{2}}{\text{SO}}_{\text{4}}\to {\text{MgSO}}_{\text{4}}{\text{+H}}_{\text{2}}$

• To get the moles of ${\text{H}}_{\text{2}}$ produced, multiply the mass of magnesium used in the reaction by its molar mass and its stoichiometric ratio with ${\text{H}}_{\text{2}}$ .

  $\text{10}\text{.00Mg}\times \frac{\text{1molMg}}{\text{24}\text{.3gMg}}\times \frac{{\text{1molH}}_{\text{2}}}{\text{1molMg}}\text{ = 0}{\text{.413molH}}_{\text{2}}$

• Recall that any gas at STP occupies a volume of $22.4L$ per mole. This can be used to get the volume of hydrogen gas produced by the reaction, by multiplying it with moles of ${\text{H}}_{\text{2}}$ produced. Just convert the units from $L$ to $c{m}^3$ since it is the unit asked by the problem.

  $\text{0}{\text{.412molH}}_{\text{2}}\times \frac{\text{22}\text{.4L}}{\text{1mol}}\times \frac{{\text{1000cm}}^{\text{3}}}{\text{1L}}\text{ = 9}{\text{.23×10}}^{\text{3}}{\text{H}}_{\text{2}}$

Conclusion

• Moles of hydrogen gas = $\text{0}{\text{.412 mol H}}_{\text{2}}$
• The hydrogen gas produced if the magnesium reacts completely = $\text{9}{\text{.23×10}}^{\text{3}}{\text{cm}}^{\text{3}}{\text{H}}_{\text{2}}$