Đáp án:

$M:\ Fe$

$m=3,6(g)$

Giải thích các bước giải:

$n_{MO}=\dfrac{m}{M+16}(mol)$

$m_{H_2SO_4}=\dfrac{156,8×5}{100}=7,84(g)$

$n_{H_2SO_4}=\dfrac{7,84}{98}=0,08(mol)$

$MO+H_2SO_4→MSO_4+H_2O$

Theo PT: $n_{H_2SO_4\ (pư)}=n_{MO}=\dfrac{m}{M+16}(g)$

$m_{dd\ X}=m+156,8\ (g)$

$m_{H_2SO_4\ dư}=7,84-98.(0,08-\dfrac{m}{M+16})=7,84-\dfrac{98m}{M+16}\ (g)$

$C\%_{H_2SO_4\ (dư)}=\dfrac{7,84-\dfrac{98m}{M+16}}{m+156,8}×100\%=1,8329\%$ $(1)$

Mặt khác:

$n_{CaCO_3}=\dfrac{5}{100}=0,05(mol)$

$CO_2+Ca(OH)_2→CaCO_3+H_2O$

Theo PT: $n_{CO_2}=n_{CaCO_3}=0,05(mol)$

$MO+CO\xrightarrow{t^o}M+CO_2$

Theo PT: $n_{MO}=n_{CO_2}=0,05(mol)$

$⇒\dfrac{m}{M+16}=0,05$ $(2)$

Từ $(1)(2)$

⇒$M=56\ (Fe)$

⇒$m=3,6(g)$

# Ephemeral