Question

A $500.0ml$sample of $0.200M$sodium phosphate is mixed with $400.0ml$ of $0.289M$ barium chloride. What is the mass of the solid produced?

Short answer

The mass of solid produced is Barium phosphate, and its mass is for the moles of $B{a}_3{\left(P{O}_4\right)}_2$

Step-by-step solution

Balancing the reactants and products

Sodium phosphate and barium chloride come together to produce barium phosphate and sodium chloride. And the equations are:$2N{a}_{3}P{O}_{4\left(aq\right)}+3BaC{l}_{2\left(aq\right)}\to B{a}_3{\left(P{O}_4\right)}_{2\left(aq\right)}+6NaC{l}_{\left(aq\right)}$

Moles of sodium phosphate

According to the given information, $500.0ml$is the volume of sodium phosphate. To calculate the moles of $N{a}_{3}P{O}_4$

$$\begin{gathered} MolesofN{a}_{3}P{O}_4=Volume\times molarity \\ =\left(500.0ml\right)\left(\frac{1.0L}{1000ml}\right)\left(\frac{0.200ml}{1.0L}\right) \\ MolesofN{a}_{3}P{O}_4=0.100molN{a}_{3}P{O}_4 \end{gathered}$$

Moles of Barium phosphate

So, by the above equation, the moles of $N{a}_{3}P{O}_4$is $0.100mol$. When $0.100mol$of $N{a}_{3}P{O}_4$ is used, Barium phosphate is produced and can be calculated as:

$$\begin{gathered} MolesofB{a}_3{\left(P{O}_4\right)}_2=\left(\frac{1molB{a}_3{\left(P{O}_4\right)}_2}{2molN{a}_{3}P{O}_4}\right)\left(0.100N{a}_{3}P{O}_4\right) \\ MolesofB{a}_3{\left(P{O}_4\right)}_2=0.05molB{a}_3{\left(P{O}_4\right)}_2 \end{gathered}$$

Moles of Barium chloride and barium phosphate.

The number of moles of $BaC{l}_2$present in $400.0ml$of$0.289MBaC{l}_2$can be calculated as:

localid="1648789985210" $$\begin{gathered} MolesofBaCl2=Volume\times Molarity \\ =\left(400.0ml\right)\times \left(\frac{1.0L}{1000ml}\right)\left(\frac{0.289mol}{1.0L}\right) \\ MolesofBaCl2=0.1156\mathrm{mol}\mathrm{BaCl}2 \end{gathered}$$

When $0.1156molofBaC{l}_2$is used, Barium phosphate is produced and can be calculated as:

$$\begin{gathered} MolesofB{a}_3{\left(P{O}_4\right)}_2=\left(\frac{1molB{a}_3{\left(P{O}_4\right)}_2}{3molBaC{l}_2}\right)\left(0.1156molBaC{l}_2\right) \\ MolofB{a}_3{\left(P{O}_4\right)}_2=0.385molB{a}_3{\left(P{O}_4\right)}_2 \end{gathered}$$

$$\begin{gathered} MolesofB{a}_3{\left(P{O}_4\right)}_2=\left(\frac{1molofB{a}_3{\left(P{O}_4\right)}_2}{3molofBaC{l}_2}\right)\left(0.1156molBaC{l}_2\right) \\ MolesofB{a}_3{\left(P{O}_4\right)}_2=0.385molofB{a}_3{\left(P{O}_4\right)}_2 \end{gathered}$$

Mass of Barium phosphate

$601.92g/mol$is the molar mass of $B{a}_3{\left(P{O}_4\right)}_2$. So the mass of $0.0385mol$ moles can be calculated as:

$$\begin{gathered} Massof0.0385molofB{a}_3{\left(P{O}_4\right)}_2=No.ofmoles\times Molarmass \\ =0.0385mol\times \left(\frac{601.92g}{mol}\right) \\ Massof0.0385molofB{a}_3{\left(P{O}_4\right)}_2=23.2g \end{gathered}$$

Conclusion

Therefore, the mass of the solid produced is Barium phosphate, and its mass is$23.2g$ for the moles of$0.0385molofB{a}_3{\left(P{O}_4\right)}_2$