Bài 12:
Giả sử thoát ra 1 mol B ($CO_2$)
$m_B=44g\Rightarrow m_A=44:44\%=100g$
$MCO_3+H_2SO_4\to MSO_4+CO_2+H_2O$
$MO+H_2SO_4\to MSO_4+H_2O$
$\Rightarrow n_{MSO_4}=n_{MCO_3}=1 mol$
$\Sigma m_{MSO_4}=100.168\%=168g$
$\Rightarrow n_{MSO_4}=\dfrac{168}{M+96}$
$\Rightarrow n_{MSO_4(+MO)}=\dfrac{168}{M+96}-1=\dfrac{72-M}{M+96}=n_{MO}$
$\Rightarrow \dfrac{72-M}{M+96}(M+16)+M+60=100$
$\Leftrightarrow M=24(Mg)$
$\%MgCO_3=\dfrac{1.84.100}{100}=84\%$
$\Rightarrow \%MO=16\%$
