Đáp án:
\(\begin{array}{l}
{C_{{M_{CuS{O_4}}}}} = 0,35M\\
{C_{{M_{FeS{O_4}}}}} = 0,35M
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Fe + CuS{O_4} \to FeS{O_4} + Cu\\
b)\\
{n_{Fe}} = \dfrac{m}{M} = \dfrac{{1,96}}{{56}} = 0,035mol\\
{m_{{\rm{dd}}CuS{O_4}}} = V \times d = 100 \times 1,12 = 112g\\
{m_{CuS{O_4}}} = \dfrac{{112 \times 10\% }}{{100}} = 11,2g\\
{n_{CuS{O_4}}} = \dfrac{m}{M} = \dfrac{{11,2}}{{160}} = 0,07mol\\
\dfrac{{0,035}}{1} < \dfrac{{0,07}}{1} \Rightarrow CuS{O_4}\text{ dư}\\
{n_{CuS{O_4}d}} = 0,07 - 0,035 = 0,035mol\\
{n_{FeS{O_4}}} = {n_{Fe}} = 0,035mol\\
{C_{{M_{CuS{O_4}}}}} = \dfrac{n}{V} = \dfrac{{0,035}}{{0,1}} = 0,35M\\
{C_{{M_{FeS{O_4}}}}} = \dfrac{n}{V} = \dfrac{{0,035}}{{0,1}} = 0,35M
\end{array}\)
