Đáp án:
\(\begin{array}{l}
a)\\
{m_{AgCl}} = 21,525g\\
b)\\
{m_{{\rm{dd}}AgN{O_3}}} = 300g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
NaCl + AgN{O_3} \to AgCl + NaN{O_3}\\
a)\\
{n_{NaCl}} = \dfrac{m}{M} = \dfrac{{8,775}}{{58,5}} = 0,15mol\\
\Rightarrow {n_{AgCl}} = {n_{NaCl}} = 0,15mol\\
{m_{AgCl}} = n \times M = 0,15 \times 143,5 = 21,525g\\
b)\\
{n_{AgN{O_3}}} = {n_{NaCl}} = 0,15mol\\
{m_{AgN{O_3}}} = n \times M = 0,15 \times 170 = 25,5g\\
{m_{{\rm{dd}}AgN{O_3}}} = \dfrac{{m \times 100\% }}{{C\% }} = \dfrac{{25,5 \times 100}}{{8,5}} = 300g
\end{array}\)
