Em tham khảo nha :
\(\begin{array}{l}
a)\\
Cu + 2{H_2}S{O_4} \to CuS{O_4} + S{O_2} + 2{H_2}O\\
{n_{S{O_2}}} = \dfrac{{1,176}}{{22,4}} = 0,0525mol\\
{n_{Cu}} = {n_{S{O_2}}} = 0,0525mol\\
{m_{Cu}} = 0,0525 \times 64 = 3,36g\\
{m_{Fe}} = 27 - 3,36 = 23,64g\\
\% Fe = \dfrac{{23,64}}{{27}} \times 100\% = 87,56\% \\
b)\\
{n_{{H_2}S{O_4}}} = 2{n_{S{O_2}}} = 0,105mol\\
{m_{{H_2}S{O_4}}} = 0,105 \times 98 = 10,29g\\
{m_{dd{H_2}S{O_4}}} = \dfrac{{10,29 \times 100}}{{80}} = 12,8625g\\
c)\\
{n_{CuS{O_4}}} = {n_{S{O_2}}} = 0,0525mol\\
{m_{CuS{O_4}}} = 0,0525 \times 160 = 8,4g\\
{m_{ddspu}} = 3,36 + 12,8625 - 0,0525 \times 64 = 12,8625g\\
C{\% _{CuS{O_4}}} = \dfrac{{8,4}}{{12,8625}} \times 100\% = 65,3\% \\
d)\\
NaOH + S{O_2} \to NaHS{O_3}\\
2NaOH + S{O_2} \to N{a_2}S{O_3} + {H_2}O\\
{C_{{M_{NaHS{O_3}}}}} = {C_{{M_{N{a_2}S{O_3}}}}} \Rightarrow {n_{NaHS{O_3}}} = {n_{N{a_2}S{O_3}}}\\
{n_{NaHS{O_3}}} + {n_{N{a_2}S{O_3}}} = {n_{S{O_2}}}\\
\Rightarrow {n_{NaHS{O_3}}} = {n_{NaOH}} = \dfrac{{0,0525}}{2} = 0,02625mol\\
{n_{NaOH}} = {n_{NaHS{O_3}}} + 2{n_{N{a_2}S{O_3}}} = 0,07875mol\\
{V_{NaOH}} = \dfrac{{0,07875}}{{0,25}} = 0,315l = 315ml\\
\end{array}\)
