Cốc 1:
Na2CO3+ 2HCl -> 2NaCl+ CO2+ H2O
nNa2CO3= $\frac{15,9}{106}$= 0,15 mol= nCO2
=> mCO2= 0,15.44= 6,6g
$\Delta m_1$= mNa2CO3-mCO2= 15,9-6,6= 9,3g
Cốc 2:
nM= $\frac{10,4625}{M}$ mol
2M+ 3H2SO4 -> M2(SO4)3+ 3H2
=> nH2= $\frac{15,69375}{M}$ mol
=> mH2= $\frac{31,3875}{M}$ g
$\Delta m_2$= mM- mH2= 10,4625- $\frac{31,3875}{M}$ g
Vì hai cân thăng bằng nên $\Delta m_1= \Delta m_2$
=> $ 9,3= 10,4625- \frac{31,3875}{M}$
<=> $M= 27$
Vậy M là Al
