Bài 6 :
a)
$Mg + H_2SO_4 \to MgSO_4 + H_2$
$2Al +3 H_2SO_4 \to Al_2(SO_4)_3 + 3H_2$
b)
Gọi $n_{Mg} = a(mol) ; n_{Al} = b(mol) \Rightarrow 24a + 27b = 15,6(1)$
THeo PTHH :
$n_{H_2} = a + 1,5b = \dfrac{17,92}{22,4} = 0,8(2)$
Từ (1)(2) suy ra a = 0,2 ; b = 0,4
$m_{Mg} = 0,2.24 = 4,8(gam)$
$m_{Al} = 0,4.27 = 10,8(gam)$
c)
$n_{H_2SO_4} = n_{H_2} = 0,8(mol)$
$V_{dd\ H_2SO_4} = \dfrac{0,8}{1,5} = 0,533(lít)$
Bài 7 :
a)
$Mg + H_2SO_4 \to MgSO_4 + H_2$
$Fe + H_2SO_4 \to FeSO_4 + H_2$
Theo PTHH :
$n_{H_2} = n_{H_2SO_4} = \dfrac{300.9,8\%}{98} = 0,3(mol)$
$V_{H_2} = 0,3.22,4 = 6,72(lít)$
$m_{H_2} = 0,3.2 = 0,6(gam)$
b) Gọi $n_{Mg} = a (mol) ; n_{Fe} = b(mol)$
Ta có :
$m_{hh} = 24a + 56b = 10,4(gam)$
$n_{H_2SO_4} = a + b = 0,3(mol)$
Suy ra a = 0,2 ; b = 0,1
$\%m_{Mg} = \dfrac{0,2.24}{10,4}.100\% = 46,15\%$
$\%m_{Fe} = 100\% -46,15\% = 53,85\%$
c)
$m_{dd\ sau\ pư} = 10,4 + 300 - 0,6 = 309,8(gam)$
$C\%_{MgSO_4} = \dfrac{0,2.120}{309,8}.100\% = 7,75\%$
$C\%_{FeSO_4} = \dfrac{0,1.152}{309,8}.100\% = 4,9\%$
