\(\rm PTHH: CuO+H_2\longrightarrow H_2O+Cu\)
a/ \(n_{CuO}=\dfrac{m_{CuO}}{M_{CuO}}=\dfrac{2,4}{80}=0,03(mol)\)
Theo pt ta có tỉ lệ: \(1:1:1:1\)
\(→n_{H_2}=0,03(mol)\)
\(→V_{H_2}=n_{H_2}.22,4=0,03.22,4=0,672(g)\)
b/ \(\rm PTHH: 2Al+3H_2SO_4\longrightarrow Al_2(SO_4)_3+3H_2)\)
Theo pt ta có tỉ lệ: \(2:3:1:3\)
\(→n_{Al}=\dfrac{2}{3}n_{H_2}=\dfrac{2}{3}.0,03=0,02(mol)\)
\(→m_{Al}=n_{Al}.M_{Al}=0,02.27=0,54(g)\)