a)4 Al+3H2SO4---->2Al2(SO4)3+3H2
n\(_{Al}=\frac{5,4}{27}=0,2\left(mol\right)\)
Theo pthh
n\(_{H2}=\frac{3}{4}n_{Al}=0,15\left(mol\right)\)
V\(_{H2}=0,15.22,4=3,36\left(l\right)\)
Theo pthh
n\(_{H2SO4}=\frac{3}{4}n_{Al}=0,15\left(mol\right)\)
m\(_{H2SO4}=0,15.98=14,7\left(g\right)\)
c) H2+ CuO ---->Cu+H2O
n\(_{C_{ }uO}=\frac{4}{80}=0,05\left(mol\right)\)
=> H2 dư
Theo pthh
n\(_{Cu}=n_{CuO}=0,05\left(mol\right)\)
m\(_{Cu}=0,05.64=3,2\left(g\right)\)
\(nAl=0,2\left(mol\right)\)
\(PTHH:4Al+3H2SO4\rightarrow2Al2\left(SO4\right)3+3H2\)
=>nH2 = 3/4 Al = 0,15 (mol)
=>VH2 = 0,15 . 22,4 = 3,36l
Ta cs : nH2SO4 = 0,14 (mol)=>mH2SO4 = 0,14 . 98 = 14,7l
\(PTHH:CuO+H2\rightarrow Cu+H2O\)
\(nCuO=0,0,5\left(mol\right)\Rightarrow nCu=nCuO=0,05\left(mol\right)\)
\(\Rightarrow mCu=0,05.64=3,2\left(g\right)\)