\(n_{CaCO_3}=\dfrac{25}{100}=0.25\left(mol\right)\)
\(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
Vì cân ở vị trí cân bằng nên :
\(m_{CaCO_3}+m_{dd_{HCl}}-m_{CO_2}=m_{Al}+m_{dd_{H_2SO_4}}-m_{H_2}\)
\(\Rightarrow m_{CaCO_3}-m_{CO_2}=m_{Al}-m_{H_2}\)
\(\Rightarrow25-0.25\cdot44=a-\dfrac{a}{27}\cdot\dfrac{3}{2}\cdot2\)
\(\Rightarrow a=15.75\left(g\right)\)
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