\(Na_2CO_3+HCl\rightarrow2NaCl+H_2O+CO_2\)
0,24_____________________________0,24
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
m/27 ________________________3m/54
\(n_{Na2CO3}=0,24\left(mol\right)\)
\(m_{Al}=\frac{m}{27}\left(mol\right)\)
\(m_{HCl}=25,44-0,24.44=14,88\left(g\right)\)
Vì đem cân nên:
\(m-\frac{3m}{54}=14,88\Rightarrow m=16,74\left(g\right)\)