The precipitation reaction is as follows:
[tex]K_{2}SO_{4}(aq)+Sr(NO_{3})_{2}(aq)\rightarrow SrSO_{4}(s)+2KNO_{3}(aq)[/tex]
Here, 1 mole of reactant 1 [tex]K_{2}SO_{4}(aq)[/tex] reacts with 1 mole of reactant 2 [tex]Sr(NO_{3})_{2}[/tex].
The mass of [tex]Sr(NO_{3})_{2}[/tex] is 56.5 g and molar mass is 211.63 g/mol, converting mass into molar mass as follows:
[tex]n=\frac{m}{M}=\frac{56.5g}{211.63 g/mol}=0.267 mol[/tex]
Since, 1 mol of [tex]Sr(NO_{3})_{2}[/tex] reacts with 1 mol of [tex]K_{2}SO_{4}(aq)[/tex] thus, 0.267 mol will react with 0.267 mol.
Molar mass of [tex]K_{2}SO_{4}(aq)[/tex] is 174.259 g/mol, mass of [tex]K_{2}SO_{4}(aq)[/tex] can be calculated as follows:
[tex]m=n\times M=0.267 mol\times 174.259 g/mol=46.52 g[/tex]
Therefore, mass of first reactant [tex]K_{2}SO_{4}(aq)[/tex] is 46.52 g
