Em tham khảo nha :
\(\begin{array}{l}
3)\\
a)\\
Fe + CuS{O_4} \to FeS{O_4} + Cu\\
b)\\
{n_{Fe}} = \dfrac{{19,6}}{{56}} = 0,35mol\\
{m_{CuS{O_4}}} = \dfrac{{300 \times 24}}{{100}} = 72g\\
{n_{CuS{O_4}}} = \dfrac{{72}}{{160}} = 0,45mol\\
\dfrac{{0,35}}{1} < \dfrac{{0,45}}{1} \Rightarrow CuS{O_4}\text{ dư}\\
{n_{Cu}} = {n_{Fe}} = 0,35mol\\
{m_{Cu}} = 0,35 \times 64 = 22,4g\\
c)\\
{m_{ddspu}} = 19,6 + 300 - 22,4 = 297,2g\\
{n_{FeS{O_4}}} = {n_{Fe}} = 0,35mol\\
{m_{FeS{O_4}}} = 0,35 \times 152 = 53,2g\\
C{\% _{FeS{O_4}}} = \dfrac{{53,2}}{{297,2}} \times 100\% = 17,9\% \\
{n_{{H_2}S{O_4}d}} = 0,45 - 0,35 = 0,1mol\\
{m_{CuS{O_4}}} = 0,1 \times 160 = 16g\\
C{\% _{CuS{O_4}}} = \dfrac{{16}}{{297,2}} \times 100\% = 5,38\% \\
4)\\
a)\\
CuO + {H_2}S{O_4} \to CuS{O_4} + {H_2}O\\
Cu + 2{H_2}S{O_4}_n \to CuS{O_4} + S{O_2} + 2{H_2}O\\
A:\text{Lưu huỳnh dioxit}\\
b)\\
{n_{S{O_2}}} = \dfrac{{1,12}}{{22,4}} = 0,05mol\\
{n_{Cu}} = {n_{S{O_2}}} = 0,05mol\\
{m_{Cu}} = 0,05 \times 64 = 3,2g\\
\% Cu = \dfrac{{3,2}}{{10}} \times 100\% = 32\% \\
\% CuO = 100 - 32 = 68\%
\end{array}\)
