$\text{Câu 6:}$
$m_{H_2SO_4}=\dfrac{245.20%}{100%}=49\text{(g)}$
$\to n_{H_2SO_4}=\dfrac{49}{98}=0,5\text{(mol)}$
$n_{H_2}=\dfrac{4,48}{22,4}=0,2\text{(mol)}$
$\text{Phương trình:}$
$Fe + H_2SO_4 \to FeSO_4 + H_2$
$\text{Do} \dfrac{0,5}{1}>\dfrac{0,2}{1}$
$\text{ Kê theo mol} H_2$
$\to n_{H_2SO_4\text{dư}}=0,3\text{(mol)}$
$\to m_{FeSO_4}=0,2.152=30,4\text{(g)}$
$\to C_{FeSO_4}=\dfrac{30,4.100}{245}=12,4$%
$m_{H_2SO_4\text{dư}}=0,3.98=29,4\text{(g)}$
$\to C_{H_2SO_4\text{dư}}=\dfrac{29,4.100}{245}=12$%
