\({\text{100.0 c}}{{\text{m}}^{\text{3}}}\) 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