Chapter 10, Problem 45E

A solution containing $1.63\text{g}$ of barium chloride is added to a solution containing $2.40\text{g}$ of sodium chromate (chromate ion, ${\text{CrO}}_4^{2-}$). Find the mass in grams of barium chromate that can precipitate. Also determine which reactant was in excess, as well as the mass in grams in excess of the amount required by the limiting reactant.

Interpretation Introduction

Interpretation:

The mass of precipitated barium chromate when barium chloride reacts with sodium chromate is to be calculated. The reactant and its mass which is present in excess in the reaction mixture is to be determined.

Concept introduction:

The limiting reagent of a reaction is that reactant of the reaction that controls the amount of product formed. The limiting agents limit the amount of product and by adding some more amount of the limiting reagent in the reaction mixture, the amount of product can be increased.

Answer to Problem 45E

The mass of precipitated barium chromate when barium chloride reacts with sodium chromate is $1.98\text{g}$. The reactant that is present in excess is sodium chromate. The mass of sodium chromate present in excess is $\text{1}\text{.13}\text{g}$.

Explanation of Solution

The reaction between barium chloride and sodium chromate is shown below.

${\text{BaCl}}_{\text{2}}+{\text{Na}}_2{\text{CrO}}_{\text{4}}\to 2\text{NaCl}+{\text{BaCrO}}_{\text{4}}$

The mass of ${\text{BaCl}}_{\text{2}}$ reacted is $1.63\text{g}$.

The mass of ${\text{Na}}_2{\text{CrO}}_{\text{4}}$ reacted is $2.40\text{g}$.

The molar mass of ${\text{BaCl}}_{\text{2}}$ is $208.2\text{g}/\text{mol}$.

The molar mass of ${\text{Na}}_2{\text{CrO}}_{\text{4}}$ is $161.98\text{g}/\text{mol}$.

The molar mass of ${\text{BaCrO}}_{\text{4}}$ is $253.3\text{g}/\text{mol}$.

The number of moles of a substance is given by the expression shown below.

$n=\frac{m}{M}$ …(1)

Where,

• $m$ is the mass of the substance.

• $M$ is the molar mass of the substance.

Substitute the mass and molar mass of ${\text{BaCl}}_{\text{2}}$ in the equation (1).

$\begin{array}{rll}n& =& \frac{1.63\text{g}}{208.2\text{g}/\text{mol}}\\ & =& 7.829\times {10}^{-3}\text{mol}\end{array}$

Therefore, the number of moles of ${\text{BaCl}}_{\text{2}}$ present in the reaction mixture is $7.829\times {10}^{-3}\text{mol}$.

Substitute the mass and molar mass of ${\text{Na}}_2{\text{CrO}}_{\text{4}}$ in the equation (1).

$\begin{array}{rll}n& =& \frac{2.40\text{g}}{161.98\text{g}/\text{mol}}\\ & =& 1.48\times {10}^{-2}\text{mol}\end{array}$

Therefore, the number of moles of ${\text{Na}}_2{\text{CrO}}_{\text{4}}$ present in the reaction mixture is $1.48\times {10}^{-2}\text{mol}$.

One mole of ${\text{BaCl}}_{\text{2}}$ reacts with one mole of ${\text{Na}}_2{\text{CrO}}_{\text{4}}$. Therefore, the relation between the number of moles of ${\text{BaCl}}_{\text{2}}$ and ${\text{Na}}_2{\text{CrO}}_{\text{4}}$ is given by the expression shown below.

$n_{{\text{BaCl}}_{\text{2}}}=n_{{\text{Na}}_2{\text{CrO}}_{\text{4}}}$ …(2)

Where,

• $n_{{\text{Na}2{{\text{CrO}\text{4}{}_{}}_{}}_{}}_{}}$ is the number of moles of ${\text{Na}}_2{\text{CrO}}_{\text{4}}$.

• $n_{{\text{BaCl}\text{2}{}_{}}_{}}$ is the number of moles of ${\text{BaCl}}_{\text{2}}$.

Substitute the value of $n_{{\text{Na}}_2{\text{CrO}}_{\text{4}}}$ in the equation (2).

$n_{{\text{BaCl}}_{\text{2}}}=1.48\times {10}^{-2}\text{mol}$

The required amount of ${\text{BaCl}}_{\text{2}}$ is greater than the available amount of ${\text{BaCl}}_{\text{2}}$. Therefore. ${\text{BaCl}}_{\text{2}}$ is the limiting reagent. The sodium chromate is present in the excess.

One mole of ${\text{BaCl}}_{\text{2}}$ produces one mole of ${\text{BaCrO}}_{\text{4}}$. Therefore, the relation between the number of moles of ${\text{BaCrO}}_{\text{4}}$ and ${\text{BaCl}}_{\text{2}}$ is given by the expression shown below.

$n_{{\text{BaCrO}}_{\text{4}}}=n_{{\text{BaCl}}_{\text{2}}}$ …(3)

Where,

• $n_{{\text{BaCl}\text{2}{}_{}}_{}}$ is the number of moles of ${\text{BaCl}}_{\text{2}}$.

• $n_{{\text{BaCrO}\text{4}{}_{}}_{}}$ is the number of moles of ${\text{BaCrO}}_{\text{4}}$.

Substitute the value of $n_{{\text{BaCl}}_{\text{2}}}$ in the equation (3).

$n_{{\text{CH}}_3{\text{CH}}_3}=7.829\times {10}^{-3}\text{mol}$

Rearrange the equation (1) for the value of $m$.

$m=nM$ …(4)

Substitute the value of molar mass and the number of moles of ${\text{BaCrO}}_{\text{4}}$ in the equation (4).

$\begin{array}{rll}m& =& \left(7.829\times {10}^{-3}\text{mol}\right)\left(253.3\text{g}/\text{mol}\right)\\ & =& 1.98\text{g}\end{array}$

Therefore, the mass of precipitated barium chromate when barium chloride reacts with sodium chromate is $1.98\text{g}$.

The number of moles of sodium chromate that will react with $7.829\times {10}^{-3}\text{mol}$ of barium chloride is $7.829\times {10}^{-3}\text{mol}$.

Substitute the value of molar mass and the number of moles of sodium chromate in the equation (4).

$\begin{array}{rll}m& =& \left(7.829\times {10}^{-3}\text{mol}\right)\left(161.98\text{g}/\text{mol}\right)\\ & =& 1.268\text{g}\\ & \approx & 1.27\text{g}\end{array}$

Therefore, the mass of sodium chromate reacted to produce $1.98\text{g}$ of barium is $1.27\text{g}$.

The excess mass of sodium chromate is given by the expression as shown below.

$m_{{\text{Na}}_2{\text{CrO}}_{\text{4}}}=m_1-m_2$

Where,

• $m_1$ is the initial mass of sodium chromate present in the reaction mixture.

• $m_2$ is the mass of sodium chromate reacted in the reaction.

Substitute the value of $m_1$ and $m_2$ in the above equation.

$\begin{array}{rll}{m}_{{\text{Na}}_2{\text{CrO}}_{\text{4}}}& =& 2.40\text{g}-1.27\text{g}\\ & =& 1\text{.13}\text{g}\end{array}$

Therefore, the mass of sodium chromate present in excess is $\text{1}\text{.13}\text{g}$.

Conclusion

The mass of precipitated barium chromate when barium chloride reacts with sodium chromate is $1.98\text{g}$. The reactant that is present in excess is sodium chromate. The mass of sodium chromate present in excess is $\text{1}\text{.13}\text{g}$.