Chapter 9, Problem 24A

Interpretation Introduction

Interpretation:

The amount of MgO formed when 1.25 g of Mg reacts with oxygen gas needs to be deduced based on the given reaction.

Concept Introduction:

• A chemical reaction is expressed as a chemical equation having reactants and products on left and right side of the reaction arrow respectively.

  $\text{Reactants }\to \text{ Products}$

• The coefficient of a balanced chemical equation i.e. the stoichiometry gives the amount of reactants and products involved in the reaction.
• Chemical equations can therefore be used to determine the amount of products formed from a known quantity of reactants.

Answer to Problem 24A

Mass of MgO = 2.08 g

Explanation of Solution

The given reaction is:

  ${\text{2Mg(s) + O}}_{\text{2}}\text{(g) }\to \text{2MgO(s)}$

Step 1: Calculate the moles of Mg present:

Mass of Mg present = 1.25 g

Atomic mass of Mg = 24 g/mol

  $\begin{array}{l}\text{Moles of Mg = }\frac{\text{Mass of Mg}}{\text{Molar Mass Mg}}\\ =\frac{\text{1}\text{.25 g}}{24\text{ g/mol}}=0.0521\text{ moles}\end{array}$

Step 2: Calculate the moles of MgO formed:

Based on the reaction stoichiometry:

2 moles of Mg will yield 2 moles of MgO i.e. they are present in a 1:1 molar ratio

Therefore, 0.0521 moles of Mg will yield 0.0521 moles of MgO

Step 3: Calculate the mass of MgO formed:

Moles of MgO formed = 0.0521

Molecular weight of MgO = 40 g/mol

  $\begin{array}{l}\text{Mass of MgO = Moles }\times \text{ Molecular weight}\\ =0.0521\text{ moles }\times \text{ 40 g/mol = 2}\text{.08 g}\end{array}$

Conclusion

Therefore, mass of MgO produced is 2.08 g.