${\text{100.0 c}}{{\text{m}}^{\text{3}}}$ of a ${\text{0.50 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}$ solution of ${\text{BaC}}{{\text{l}}_{\text{2}}}$ is added to ${\text{50.0 c}}{{\text{m}}^{\text{3}}}$ of a ${\text{0.10 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}$ solution of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$. A precipitate of ${\text{BaS}}{{\text{O}}_{\text{4}}}$ is formed according to the equation below.

$${\text{BaC}}{{\text{l}}_2}{\text{(aq)}} + {\text{N}}{{\text{a}}_2}{\text{S}}{{\text{O}}_4}{\text{(aq)}} \to {\text{BaS}}{{\text{O}}_4}{\text{(s)}} + {\text{2NaCl(aq)}}$$

What is the amount, in mol, of ${\text{BaS}}{{\text{O}}_{\text{4}}}$ produced?

A.     0.0050

B.     0.010

C.     0.050

D.     0.10

A