Câu 356:
Ta có: 56nFe + 64nCu + 108nAg = 54,28 (1)
BTNT Fe: nFe2(SO4)3 = 1/2nFe
BTNT Cu: nCuSO4 = nCu
⇒ 200nFe - 160nCu = 27,2 (2)
\(m_{H_2SO_4}=179.98\%=175,42\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{175,42}{98}=1,79\left(mol\right)\)
Có: nH2SO4 = 2nSO2 ⇒ nSO2 = 0,895 (mol)
BT e, có: 3nFe + 2nCu + nAg = 2nSO2 = 1,79 (3)
Từ (1), (2) và (3) \(\Rightarrow\left\{{}\begin{matrix}n_{Fe}=0,36\left(mol\right)\\n_{Cu}=0,28\left(mol\right)\\n_{Ag}=0,15\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\%m_{Ag}=\dfrac{0,15.108}{54,28}.100\%\approx29,85\%\)
→ Đáp án: C
Câu 357:
Ta có: \(65n_{Zn}+56n_{Fe}+64n_{Cu}=64,34\left(1\right)\)
\(n_{H_2}=\dfrac{15,008}{22,4}=0,67\left(mol\right)\)
BT e, có: \(2n_{Zn}+2n_{Fe}=2n_{H_2}=1,34\left(2\right)\)
\(m_{H_2SO_4}=231.98\%=226,38\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{226,38}{98}=2,31\left(mol\right)\)
Mà: nH2SO4 = 2nSO2 ⇒ nSO2 = 1,155 (mol)
BT e, có: 2nZn + 3nFe + 2nCu = 2nSO2 = 2,31 (3)
Từ (1), (2) và (3) \(\Rightarrow\left\{{}\begin{matrix}n_{Zn}=0,42\left(mol\right)\\n_{Fe}=0,25\left(mol\right)\\n_{Cu}=0,36\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\%m_{Zn}=\dfrac{0,42.65}{64,34}.100\%\approx42,43\%\)
→ Đáp án: C
