a)
$M + 2HCl \to MCl_2 + H_2$
$MCO_3 + 2HCl \to MCl_2 + CO_2 + H_2O$
b)
Gọi $n_{H_2} = a(mol) ; n_{CO_2 } = b(mol)$
Ta có :
$n_Y = a + b = \dfrac{1,12}{22,4} = 0,05(mol)$
\(M_Y=\dfrac{2a+44b}{a+b}=32.0,325=10,4\)(g/mol)
Suy ra : a = 0,04 ; b = 0,01
$\%V_{H_2} = \dfrac{0,04}{0,05}.100\% = 80\%$
$\%V_{CO_2} = 100\% - 80\% = 20\%$
c)Theo PTHH :$n_{MCl_2} = n_{H_2} + n_{CO_2} = 0,05(mol)$
$\Rightarrow M_{MCl_2} = M + 71 = \dfrac{6,8}{0,05} = 136 \Rightarrow M = 65$
$m = 0,04.65 + 0,01.125 = 3,85(gam)$
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