\(n_{Al}=\dfrac{0,54}{27}=0,02\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{0,98}{98}=0,01\left(mol\right)\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)
0,02 0,01 ---------------------------> 0,01
Lập tỉ số: \(n_{Al}:n_{H_2SO_4}=\dfrac{1}{100}>\dfrac{1}{300}\)
=> Al dư, H2SO4 hết
\(V_{H_2}=0,01.22,4=0,224\left(l\right)\)
\(m_{Al\left(dư\right)}=\left(0,02-\dfrac{1}{150}\right).27=0,36\left(g\right)\)
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