Đáp án + Giải thích các bước giải:

Câu 3:

$a\big)$

$n_{CuO}=\dfrac{50.40\%}{80}=0,25(mol)$

$n_{Fe_3O_4}=\dfrac{50.60\%}{232}\approx 0,13(mol)$

$CuO+H_2\xrightarrow{t^o}Cu+H_2O$

$Fe_3O_4+4H_2\xrightarrow{t^o}3Fe+4H_2O$

Theo PT: $n_{H_2}=0,25+0,13.4=0,77(mol)$

$\to V_{H_2}=0,77.22,4=17,248(l)$

$b\big)$

Theo PT: $n_{Cu}=0,25(mol);n_{Fe}=0,39(mol)$

$\to m_{Cu}=0,25.64=16(g)$

$\to m_{Fe}=0,39.56=21,84(g)$

Câu 4:

$a\big)$

$CuO+CO\xrightarrow{t^o}Cu+CO_2$

Theo PT: $n_{CuO}=n_{CO}=\dfrac{3,36}{22,4}=0,15(mol)$

$\to \%m_{CuO}=\dfrac{0,15.80}{17,6}.100\%\approx 68,18\%$

$\to \%m_{Fe}=100-68,18=31,82\%$

$b\big)$

$n_{Fe}=\dfrac{17,6-0,15.80}{56}=0,1(mol)$

Theo PT: $n_{Cu}=n_{CO}=0,15(mol)$

$3Fe+2O_2\xrightarrow{t^o}Fe_3O_4$

$2Cu+O_2\xrightarrow{t^o}2CuO$

Theo PT: $n_{O_2}=\dfrac{2}{3}.0,1+\dfrac{1}{2}.0,15=\dfrac{17}{120}(mol)$

$\to V_{O_2}=\dfrac{17}{120}.22,4\approx 3,173(l)$