Chapter 11, Problem 5STP

Interpretation Introduction

Interpretation:

The amount of ${\text{NaH}}_{\text{2}}{\text{PO}}_{\text{4}}$ required to make 444.0 g of ${\text{Na}}_2{\text{H}}_{\text{2}}{\text{P}}_2{\text{O}}_7$ under given conditions should be determined.

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 5STP

The correct option is (c), $\text{130 g}$.

Explanation of Solution

Given:

The chemical reaction:

${\text{2NaH}}_{\text{2}}{\text{PO}}_4\text{(s) }\to {\text{ Na}}_2{\text{H}}_{\text{2}}{\text{P}}_2{\text{O}}_7{\text{(s) + H}}_{\text{2}}\text{O(g)}$

The available mass of ${\text{NaH}}_{\text{2}}{\text{PO}}_4\text{ = 350}\text{.0 g}$.

The number of moles of ${\text{Na}}_2{\text{H}}_{\text{2}}{\text{P}}_2{\text{O}}_7$ is calculated as:

$\text{Number of moles}=\frac{\text{Mass}}{\text{Molar mass}}$

The molar mass of ${\text{Na}}_2{\text{H}}_{\text{2}}{\text{P}}_2{\text{O}}_7$ is 221.94 g/mol. So, the number of moles of ${\text{Na}}_2{\text{H}}_{\text{2}}{\text{P}}_2{\text{O}}_7$ is:

$\begin{array}{l}{\text{n}}_{{\text{KClO}}_{\text{3}}}=\frac{444.0\text{ g}}{\text{221}\text{.94 g/mol}}\\ {\text{n}}_{{\text{KClO}}_{\text{3}}}=2\text{ mol}\end{array}$

From balanced reaction, the mole ratio of ${\text{Na}}_2{\text{H}}_{\text{2}}{\text{P}}_2{\text{O}}_7$ and ${\text{NaH}}_{\text{2}}{\text{PO}}_4$ is 1:2 that means 1 mole of ${\text{Na}}_2{\text{H}}_{\text{2}}{\text{P}}_2{\text{O}}_7$ is produced from 2 moles of ${\text{NaH}}_{\text{2}}{\text{PO}}_4$. So, the number of moles of ${\text{NaH}}_{\text{2}}{\text{PO}}_4$ produced from 2.0 mol of ${\text{Na}}_2{\text{H}}_{\text{2}}{\text{P}}_2{\text{O}}_7$ is:

${\text{2 mol of Na}}_2{\text{H}}_{\text{2}}{\text{P}}_2{\text{O}}_7\text{ ×}\frac{{\text{2 mol of NaH}}_{\text{2}}{\text{PO}}_4}{{\text{1 mol of Na}}_2{\text{H}}_{\text{2}}{\text{P}}_2{\text{O}}_7}{\text{ = 4 mol of NaH}}_{\text{2}}{\text{PO}}_4$

Now, the required mass of ${\text{NaH}}_{\text{2}}{\text{PO}}_4$ is calculated as:

$\text{Number of moles}=\frac{\text{Mass}}{\text{Molar mass}}$

Molar mass of ${\text{NaH}}_{\text{2}}{\text{PO}}_4$ is 119.98 g/mol. So,

$\begin{array}{l}\text{4 mol}=\frac{\text{Mass}}{\text{119}\text{.98 g/mol}}\\ \text{Mass = 4 mol}\times \text{119}\text{.98 g/mol}\\ \text{Mass = 479}\text{.92 g}\end{array}$

Since the available amount of ${\text{NaH}}_{\text{2}}{\text{PO}}_4\text{ = 350}\text{.0 g}$ and the required amount of ${\text{NaH}}_{\text{2}}{\text{PO}}_4\text{ = 479}\text{.9 g}$ so, the amount of ${\text{NaH}}_{\text{2}}{\text{PO}}_4$ that should be buy to make enough ${\text{Na}}_2{\text{H}}_{\text{2}}{\text{P}}_2{\text{O}}_7$ is:

$\text{479}\text{.9 g - 350 g = 129}\text{.92 g }\simeq \text{ 130 g}$

Hence, $\text{130 g}$ is the amount of ${\text{NaH}}_{\text{2}}{\text{PO}}_4$ that should be buy to make enough ${\text{Na}}_2{\text{H}}_{\text{2}}{\text{P}}_2{\text{O}}_7$.