Chapter 9, Problem 57A

Interpretation Introduction

Interpretation: The number of moles of $\text{Si}$ produced should be determined if 3.75 moles of $\text{NaF}$ are produced in the given reaction.

Concept introduction: The ratio of mass of substance to its molar mass is said to be number of moles of that substance.

The ratio of amount of any two compounds involved in a chemical reaction in terms of mole is said to be mole ratio, and used as a conversion factor between reactants and products.

Answer to Problem 57A

  $0.625\text{ mol of Si}$ .

Explanation of Solution

Given:

  ${\text{Na}}_{\text{2}}{\text{SiF}}_{\text{6}}\left(s\right)\text{ + Na}\left(s\right)\to \text{ Si}\left(s\right)\text{ + NaF}\left(s\right)$

The given reaction is unbalanced as the number of atom of Na and F on the reactant side is 3 and 6, whereas on the product side is 1 so, in order to balance the reaction the coefficient 4 is written before Na on the reactant side and coefficient 6 is written before NaF on the product side. So, the balanced reaction is:

  ${\text{Na}}_{\text{2}}{\text{SiF}}_{\text{6}}\left(s\right)\text{ + 4Na}\left(s\right)\to \text{ Si}\left(s\right)\text{ + 6NaF}\left(s\right)$

From the balanced chemical reaction, the mole ratio of $\text{Si}$ and $\text{NaF}$ is 1:6 that means for every 1 mole of $\text{Si}$ produced there is 6 moles of $\text{NaF}$ formed. So, the number of moles of $\text{Si}$ produced from $3.75\text{ mol}$ of $\text{NaF}$ is:

  $3.75\text{ mol of NaF×}\frac{\text{1 mol of Si}}{\text{6 mol of NaF}}\text{ = }0.625\text{ mol of Si}$

Interpretation Introduction

Interpretation: The mass of ${\text{Na}}_{\text{2}}{\text{SiF}}_{\text{6}}$ (in grams) required to react with $0.555\text{ mol}$ of $\text{Na}$ needs to be calculated.

Concept introduction: The ratio of mass of substance to its molar mass is said to be number of moles of that substance.

The ratio of amount of any two compounds involved in a chemical reaction in terms of mole is said to be mole ratio, and used as a conversion factor between reactants and products.

Answer to Problem 57A

The mass of ${\text{Na}}_{\text{2}}{\text{SiF}}_{\text{6}}$ required to react with $0.555\text{ mol}$ of $\text{Na}$ is $26.14\text{ g}$ .

Explanation of Solution

The balanced reaction is:

  ${\text{Na}}_{\text{2}}{\text{SiF}}_{\text{6}}\left(s\right)\text{ + 4Na}\left(s\right)\to \text{ Si}\left(s\right)\text{ + 6NaF}\left(s\right)$

From the balanced chemical reaction, the mole ratio of ${\text{Na}}_{\text{2}}{\text{SiF}}_{\text{6}}$ and $\text{Na}$ is 1:4 that means 1 mole of ${\text{Na}}_{\text{2}}{\text{SiF}}_{\text{6}}$ reacts with 4 moles of $\text{Na}$ . So, the number of moles of ${\text{Na}}_{\text{2}}{\text{SiF}}_{\text{6}}$ produced from $0.555\text{ mol}$ of $\text{Na}$ is:

  $0.555\text{ mol of Na×}\frac{{\text{1 mol of Na}}_{\text{2}}{\text{SiF}}_{\text{6}}}{\text{4 mol of Na}}\text{ = }0.139{\text{ mol of Na}}_{\text{2}}{\text{SiF}}_{\text{6}}$

Now, the mass of ${\text{Na}}_{\text{2}}{\text{SiF}}_{\text{6}}$ reacted with $0.555\text{ mol}$ of $\text{Na}$ is calculated as:

The molar mass of ${\text{Na}}_{\text{2}}{\text{SiF}}_{\text{6}}$ is 188.05 g/mol. So,

  $\begin{array}{l}\text{Number of moles}=\frac{\text{Mass}}{\text{Molar mass}}\\ 0.139\text{ mol }=\frac{\text{Mass}}{188.05\text{ g/mol}}\\ \text{Mass = }0.139\text{ mol }\times 188.05\text{ g/mol}\\ \text{Mass = }26.14\text{ g}\end{array}$

Hence, the mass of ${\text{Na}}_{\text{2}}{\text{SiF}}_{\text{6}}$ required to react with $0.555\text{ mol}$ of $\text{Na}$ is $26.14\text{ g}$ .