Chapter 7, Problem 7.74AP

(a)

Interpretation Introduction

Interpretation:

The grams of sodium nitrate present have to be calculated.

Concept Introduction:

Molarity:  Molarity is defined as the mass of solute in one liter of solution.  Molarity is the preferred concentration unit for stoichiometry calculations.  The formula is,

  $\text{Molarity=}\frac{\text{Mass}\text{of}\text{solute(in}\text{moles)}}{\text{Volume}\text{of}\text{solution}\text{(in}\text{litres)}}$

Answer to Problem 7.74AP

The grams of sodium nitrate is $\text{12g}\text{.}$

Explanation of Solution

Given,

Volume of solution=$\text{250mL}$

Molarity of solution=$\text{0}\text{.55M}$

Milliliters is converted into liters,

  $\begin{array}{l}\text{L=250mL×}\frac{\text{1L}}{\text{1000mL}}\\ \text{L=0}\text{.250L}\end{array}$

The moles of sodium nitrate is calculated as,

  $\begin{array}{l}\text{Molarity=}\frac{\text{Mass}\text{of}\text{solute(in}\text{moles)}}{\text{Volume}\text{of}\text{solution}\text{(in}\text{litres)}}\\ \text{0}\text{.55M=}\frac{\text{x}}{\text{0}\text{.250L}}\\ \text{x=}\left(\text{0}\text{.250L}\right)\left(0.55\frac{\text{mol}}{\text{L}}\right)\\ \text{x=0}\text{.1375mole}\end{array}$

The moles of sodium nitrate is $\text{0}\text{.14mol}\text{.}$

Moles to grams is converted as,

  $\begin{array}{l}\text{Grams=0}{\text{.14mol NaNO}}_{\text{3}}\text{×}\frac{\text{85}{\text{.00g NaNO}}_{\text{3}}}{\text{1mole}{\text{NaNO}}_{\text{3}}}\\ \text{Grams=11}\text{.9g}\end{array}$

The grams of sodium nitrate is $\text{12g}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The grams of nitric acid present have to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 7.74AP

The grams of nitric acid is $\text{37g}\text{.}$

Explanation of Solution

Given,

Volume of solution=$\text{145mL}$

Molarity of solution=$\text{4}\text{.0M}$

Milliliters is converted into liters,

  $\begin{array}{l}\text{L=145mL×}\frac{\text{1L}}{\text{1000mL}}\\ \text{L=0}\text{.145L}\end{array}$

The moles of nitric acid is calculated as,

  $\begin{array}{l}\text{Molarity=}\frac{\text{Mass}\text{of}\text{solute(in}\text{moles)}}{\text{Volume}\text{of}\text{solution}\text{(in}\text{litres)}}\\ \text{4}\text{.0M=}\frac{\text{x}}{\text{0}\text{.145L}}\\ \text{x=}\left(\text{0}\text{.145L}\right)\left(4.0\frac{\text{mol}}{\text{L}}\right)\\ \text{x=0}\text{.58mole}\end{array}$

The moles of nitric acid is $\text{0}\text{.58mol}\text{.}$

Moles to grams is converted as,

  $\begin{array}{l}\text{Grams=0}{\text{.58mol HNO}}_3\text{×}\frac{\text{63}{\text{.02g HNO}}_3}{\text{1mole}{\text{HNO}}_3}\\ \text{Grams=36}\text{.5516g}\end{array}$

The grams of nitric acid is $\text{37g}\text{.}$

(c)

Interpretation Introduction

Interpretation:

The grams of hydrochloric acid present have to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 7.74AP

The grams of hydrochloric acid is $\text{583}\text{.36g}\text{.}$

Explanation of Solution

Given,

Volume of solution=$\text{6}\text{.5L}$

Molarity of solution=$\text{2}\text{.5M}$

The moles of hydrochloric acid is calculated as,

  $\begin{array}{l}\text{Molarity=}\frac{\text{Mass}\text{of}\text{solute(in}\text{moles)}}{\text{Volume}\text{of}\text{solution}\text{(in}\text{litres)}}\\ \text{2}\text{.5M=}\frac{\text{x}}{\text{6}\text{.5L}}\\ \text{x=}\left(\text{6}\text{.5L}\right)\left(\text{2}\text{.5}\frac{\text{mol}}{\text{L}}\right)\\ \text{x=16}\text{.25mole}\end{array}$

The moles of hydrochloric acid is $\text{16mol}\text{.}$

Moles to grams is converted as,

  $\begin{array}{l}\text{Grams=16mol HCl×}\frac{\text{36}\text{.46g HCl}}{\text{1mole}\text{HCl}}\\ \text{Grams=583}\text{.36g}\end{array}$

The grams of hydrochloric acid is $\text{583}\text{.36g}\text{.}$