Question

Answer 1

a) $2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2$
$Mg + H_2SO_4 \to MgSO_4 + H_2$

b) Gọi $n_{Al} = a(mol) ; n_{Mg} = b(mol) \Rightarrow 27a + 24b = 10,05(1)$

Theo PTHH : 

$n_{H_2} = 1,5a + b = \dfrac{10,64}{22,4} = 0,475(2)$

Từ (1)(2) suy ra : a = 0,15 ; b = 0,25

$\%m_{Al} = \dfrac{0,15.27}{10,05}.100\% = 40,3\%$
$\%m_{Fe} = 100\% - 40,3\% = 59,7\%$

c)$n_{H_2SO_4} = n_{H_2} = 0,475(mol)$
$m_{dd\ H_2SO_4} = \dfrac{0,475.98}{9,8\%} = 475(gam)$

$m_{dd\ sau\ pư} = 10,05 + 475 - 0,475.2 = 484,1(gam)$

$C\%_{Al_2(SO_4)_3} = \dfrac{0,075.342}{484,1}.100\% = 5,3\%$

$C\%_{MgSO_4} = \dfrac{0,25.120}{484,1}.100\% = 6,2\%$