Chapter 14, Problem 35PE

(a)

Interpretation Introduction

Interpretation:

Moles of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ produced from $2.7{\text{ mol Na}}_3{\text{PO}}_4$ have to be determined.

Concept Introduction:

Stoichiometry describes quantitative relationships between reactants and products in any chemical reactions. In these problems, amount of one species is determined with help of known amount of another species.

Explanation of Solution

Given reaction occurs as follows:

  $3\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\left(aq\right)+2{\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}\left(aq\right)\to {\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\left(aq\right)+6{\text{NaNO}}_{\text{3}}$

According to balanced chemical equation, three moles of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ react with two moles of ${\text{Na}}_3{\text{PO}}_4$ and produce one mole of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ and six moles of ${\text{NaNO}}_{\text{3}}$. Since stoichiometric ratio between ${\text{Na}}_3{\text{PO}}_4$ and ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ is $2:1$, amount of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ produced from 2.7 moles of ${\text{Na}}_3{\text{PO}}_4$ is calculated as follows:

  $\begin{array}{c}{\text{Amount of Ca}}_3{\left({\text{PO}}_4\right)}_2=\left(\frac{1{\text{ mol Ca}}_3{\left({\text{PO}}_4\right)}_2}{{\text{2 mol Na}}_3{\text{PO}}_4}\right)\left(\text{2}{\text{.7 mol Na}}_3{\text{PO}}_4\right)\\ =1.35\text{ mol}\end{array}$

Hence, amount of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ is $1.35\text{ mol}$.

(b)

Interpretation Introduction

Interpretation:

Moles of ${\text{NaNO}}_{\text{3}}$ produced from $0.75\text{ mol Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ have to be determined.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Given reaction occurs as follows:

  $3\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\left(aq\right)+2{\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}\left(aq\right)\to {\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\left(aq\right)+6{\text{NaNO}}_{\text{3}}$

According to balanced chemical equation, three moles of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ react with two moles of ${\text{Na}}_3{\text{PO}}_4$ and produce one mole of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ and six moles of ${\text{NaNO}}_{\text{3}}$. Since stoichiometric ratio between $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ and ${\text{NaNO}}_{\text{3}}$ is $3:6$, amount of ${\text{NaNO}}_{\text{3}}$ produced from 0.75 moles of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is calculated as follows:

  $\begin{array}{c}{\text{Amount of NaNO}}_{\text{3}}=\left(\frac{6{\text{ mol NaNO}}_{\text{3}}}{\text{3 mol Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}\right)\left(\text{0}\text{.75 mol Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\right)\\ =1.5\text{ mol}\end{array}$

Hence, amount of ${\text{NaNO}}_{\text{3}}$ is $1.5\text{ mol}$.

(c)

Interpretation Introduction

Interpretation:

Moles of ${\text{Na}}_3{\text{PO}}_4$ required to react with $1.45\text{ L}$ of $\text{0}\text{.225 }M\text{ Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ have to be determined.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Given reaction occurs as follows:

  $3\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\left(aq\right)+2{\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}\left(aq\right)\to {\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\left(aq\right)+6{\text{NaNO}}_{\text{3}}$

Expression to calculate molarity of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is as follows:

  $\text{Molarity of Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}=\frac{\text{Moles of Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}{\text{Volume }\left(\text{L}\right)\text{ of Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\text{ solution}}$        (1)

Rearrange equation (1) for moles of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$.

  $\text{Moles of Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}=\left[\begin{array}{c}\left(\text{Molarity of Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\right)\\ \left(\text{Volume }\left(\text{L}\right)\text{ of Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\text{ solution}\right)\end{array}\right]$        (2)

Substitute $\text{0}\text{.225 }M$ for molarity and $1.45\text{ L}$ for volume of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ solution in equation (2).

  $\begin{array}{c}\text{Moles of Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}=\left(\text{0}\text{.225 }M\right)\left(\text{1}\text{.45 L}\right)\\ =0.326\text{ mol}\end{array}$

According to balanced chemical equation, three moles of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ react with two moles of ${\text{Na}}_3{\text{PO}}_4$ and produce one mole of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ and six moles of ${\text{NaNO}}_{\text{3}}$. Since stoichiometric ratio between $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ and ${\text{Na}}_3{\text{PO}}_4$ is $3:2$, amount of ${\text{Na}}_3{\text{PO}}_4$ required to react with 0.326 moles of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is calculated as follows:

  $\begin{array}{c}{\text{Amount of Na}}_3{\text{PO}}_4=\left(\frac{2{\text{ mol Na}}_3{\text{PO}}_4}{\text{3 mol Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}\right)\left(\text{0}\text{.326 mol Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\right)\\ =0.217\text{ mol}\end{array}$

Hence, amount of ${\text{Na}}_3{\text{PO}}_4$ required is $0.217\text{ mol}$.

(d)

Interpretation Introduction

Interpretation:

Grams of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ obtained from $\text{125 mL}$ of $\text{0}\text{.500 }M\text{ Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ have to be determined.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Given reaction occurs as follows:

  $3\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\left(aq\right)+2{\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}\left(aq\right)\to {\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\left(aq\right)+6{\text{NaNO}}_{\text{3}}$

Expression to calculate molarity of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is as follows:

  $\text{Molarity of Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}=\frac{\text{Moles of Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}{\text{Volume }\left(\text{L}\right)\text{ of Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\text{ solution}}$        (1)

Rearrange equation (1) for moles of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$.

  $\text{Moles of Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}=\left[\begin{array}{c}\left(\text{Molarity of Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\right)\\ \left(\text{Volume }\left(\text{L}\right)\text{ of Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\text{ solution}\right)\end{array}\right]$        (2)

Substitute $\text{0}\text{.500 }M$ for molarity and $\text{125 mL}$ for volume of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ solution in equation (2).

  $\begin{array}{c}\text{Moles of Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}=\left(\text{0}\text{.500 }M\right)\left(\text{125 mL}\right)\left(\frac{{10}^{-3}\text{ L}}{1\text{ mL}}\right)\\ =0.0625\text{ mol}\end{array}$

According to balanced chemical equation, three moles of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ react with two moles of ${\text{Na}}_3{\text{PO}}_4$ and produce one mole of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ and six moles of ${\text{NaNO}}_{\text{3}}$. Since stoichiometric ratio between $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ and ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ is $3:1$, amount of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ required to react with 0.0625 moles of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is calculated as follows:

  $\begin{array}{c}{\text{Amount of Ca}}_3{\left({\text{PO}}_4\right)}_2=\left(\frac{1{\text{ mol Ca}}_3{\left({\text{PO}}_4\right)}_2}{\text{3 mol Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}\right)\left(0.0625\text{ mol Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\right)\\ =0.0208\text{ mol}\end{array}$

Expression for mass of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ is as follows:

  ${\text{Mass of Ca}}_3{\left({\text{PO}}_4\right)}_2=\left[\begin{array}{c}\left({\text{Moles of Ca}}_3{\left({\text{PO}}_4\right)}_2\right)\\ \left({\text{Molar mass of Ca}}_3{\left({\text{PO}}_4\right)}_2\right)\end{array}\right]$        (3)

Substitute $0.0208\text{ mol}$ for moles and $310.2\text{ g/mol}$ for molar mass of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ in equation (3).

  $\begin{array}{c}{\text{Mass of Ca}}_3{\left({\text{PO}}_4\right)}_2=\left(0.0208\text{ mol}\right)\left(310.2\text{ g/mol}\right)\\ =6.45\text{ g}\end{array}$

Hence, $6.45\text{ g}$ of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ is required.

(e)

Interpretation Introduction

Interpretation:

Volume of $\text{0}\text{.25 }M{\text{ Na}}_3{\text{PO}}_4$ required to react with $\text{15}\text{.0 mL}$ of $\text{0}\text{.50 }M\text{ Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ have to be determined.

Concept Introduction:

Expression for molarity equation is as follows:

  $M_1{V}_1=M_2{V}_2$        (4)

Here,

$M_1$ is molarity of one solution.

$V_1$ is volume of one solution.

$M_2$ is molarity of another solution.

$V_2$ is volume of another solution.

Explanation of Solution

Rearrange equation (4) for $V_2$.

  $V_2=\frac{M_1{V}_1}{M_2}$        (5)

Substitute $\text{0}\text{.25 }M$ for $M_1$, $\text{15}\text{.0 mL}$ for $V_1$ and $\text{0}\text{.50 }M$ for $M_2$ in equation (5) for volume of ${\text{Na}}_3{\text{PO}}_4$.

  $\begin{array}{c}{\text{Volume of Na}}_3{\text{PO}}_4=\frac{\left(\text{0}\text{.25 }M\right)\left(\text{15}\text{.0 mL}\right)}{\text{0}\text{.50 }M}\\ =7.5\text{ mL}\end{array}$

Hence volume of ${\text{Na}}_3{\text{PO}}_4$ required is $7.5\text{ mL}$.

(f)

Interpretation Introduction

Interpretation:

Molarity of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ formed by reaction of $\text{50}\text{.0 mL}$ of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ with $\text{50}\text{.0 mL}$ of $\text{2}\text{.0 }M{\text{ Na}}_3{\text{PO}}_4$ have to be determined.

Concept Introduction:

Refer to part (e).

Explanation of Solution

Rearrange equation (4) for $M_2$.

  $M_2=\frac{M_1{V}_1}{V_2}$        (6)

Substitute $\text{2}\text{.0 }M$ for $M_1$, $\text{50}\text{.0 mL}$ for $V_1$ and $\text{50}\text{.0 mL}$ for $V_2$ in equation (5) for molarity of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$.

  $\begin{array}{c}\text{Molarity of Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}=\frac{\left(\text{2}\text{.0 }M\right)\left(\text{50}\text{.0 mL}\right)}{\text{50}\text{.0 mL}}\\ =\text{2}\text{.0 }M\end{array}$

Hence molarity of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ required is $2.0M$.