Đáp án:
Giải thích các bước giải:
Gọi $m_{H_2SO_4(pư)} = a(gam)$
$⇒ m_{H_2SO_4(dư)} = a.10\%(gam)$
$⇒m_{H_2SO_4(\text{ban đầu})} = a + a.10\% = 220.49\% = 107,8$
$⇒ a = 98(gam)$
Suy ra :
$m_{H_2SO_4(pư)} = 98(gam)$
$⇒ n_{H_2SO_4} = \dfrac{98}{98} = 1(mol)$
$m_{H_2SO_4(dư)} = a.10\% = 98.10\% = 9,8(gam)$
$Fe_2O_3 + 3H_2SO_4 → Fe_2(SO_4)_3 + 3H_2O$
Theo PTHH:
$n_{Fe_2O_3} = n_{Fe_2(SO_4)_3} = \dfrac{n_{H_2SO_4}}{3} = \dfrac{1}{3}(mol)$
$⇒ m_{Fe_2O_3} = \dfrac{1}{3}.160 = 53,33(gam)$
$⇒ m_{Fe_2(SO_4)_3} = \dfrac{1}{3}.400 = 133,33(gam)$
Sau phản ứng,
$m_{dd} = m_{Fe_2O_3} + m_{\text{dd H2SO4}} = 53,33 + 220 = 273,33(gam)$
Vậy :
$C\%_{H_2SO_4(dư)} = \dfrac{9,8}{273,33}.100\$ = 3,59\%$
$C\%_{Fe_2(SO_4)_3} = \dfrac{133,33}{273,33}.100\% = 48,78\%$
