Đáp án:
\({m_{FeC{O_3}}} = 6,96g\)
\({m_{C{u_2}S}} = 12g\)
\({m_{A{g_2}S}} = 11,16g\)
Giải thích các bước giải:
Gọi a,b,c là số mol của \(FeC{O_3},C{u_2}S,A{g_2}S\)
\(\begin{array}{l}
2FeC{O_3} + \frac{1}{2}{O_2} \to F{e_2}{O_3} + 2C{O_2}\\
C{u_2}S + 2{O_2} \to 2CuO + S{O_2}\\
A{g_2}S + {O_2} \to 2Ag + S{O_2}\\
S{O_2} + B{r_2} + 2{H_2}O \to {H_2}S{O_4} + 2HBr
\end{array}\)
Khí G gồm \(C{O_2}\) và \(S{O_2}\)
Chất rắn E gồm: \(F{e_2}{O_3}\), CuO,Ag
Dung dịch F gồm: \({H_2}S{O_4},HBr\)
Kết tủa gồm: \(Fe{(OH)_3},Cu{(OH)_2},Ag,BaS{O_4}\)
\(\begin{array}{l}
{n_{S{O_2}}} = {n_{C{u_2}S}} + {n_{A{g_2}S}} = b + c(mol)\\
{n_{F{e_2}{O_3}}} = \dfrac{1}{2}{n_{FeC{O_3}}} = 0,5amol\\
{n_{CuO}} = 2{n_{C{u_2}S}} = 2bmol\\
{n_{Ag}} = 2{n_{A{g_2}S}} = 2cmol\\
{n_{{H_2}S{O_4}}} = b + c(mol)\\
{n_{HBr}} = 2b + 2c(mol)\\
F{e_2}{O_3} + 3{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3{H_2}O\\
F{e_2}{O_3} + 6HBr \to 2FeB{r_3} + 3{H_2}O\\
CuO + {H_2}S{O_4} \to CuS{O_4} + {H_2}O\\
CuO + 2HBr \to CuB{r_2} + {H_2}O\\
{n_{{H_2}O}} = {n_O} \to 2b + 2c = \dfrac{1}{2}a \times 3 + 2b\\
\to 1,5a - 2c = 0\\
Fe{(OH)_3},Cu{(OH)_2},Ag,BaS{O_4}\\
{n_{Fe{{(OH)}_3}}} = amol\\
{n_{Cu{{(OH)}_2}}} = 2bmol\\
{n_{Ag}} = 2cmol\\
{n_{BaS{O_4}}} = b + c(mol)\\
107a + 98 \times 2b + 108 \times 2c + 233(b + c) = 588,8\\
\to 107a + 429b + 449c = 58,8\\
\left\{ \begin{array}{l}
116a + 160b + 248c = 30,12\\
1,5a - 2c = 0\\
107a + 429b + 449c = 58,8
\end{array} \right.\\
\to a = 0,06 \to b = 0,075 \to c = 0,045\\
\to {m_{FeC{O_3}}} = 6,96g\\
\to {m_{C{u_2}S}} = 12g\\
\to {m_{A{g_2}S}} = 11,16g
\end{array}\)
