Question

Answer 1

a) Gọi $n_{Mg} = a(mol) ; n_{Zn} = b(mol) \Rightarrow 24a + 65b = 4,2(1)$
$Mg + 2HCl \to MgCl_2 + H_2$
$Zn + 2HCl \to ZnCl_2 + H_2$
$n_{H_2} = a + b = \dfrac{2,24}{22,4} = 0,1(2)$
Từ (1)(2) suy ra $a = \dfrac{23}{410} ; b = \dfrac{9}{205}$

$\%m_{Mg} = \(\dfrac{\dfrac{23}{410}.24}{4,2}.100\%=32,06\%\)$

$\%m_{Zn} = 100\% - 32,06\% = 67,94\%$

b)

$m_{MgCl_2} = \dfrac{23}{410}.95 = 5,33(gam)$
$m_{ZnCl_2} = \dfrac{9}{205}.136 = 5,97(gam)$