Question

Answer 1

\(n_{Al}=a\left(mol\right),n_{Fe}=b\left(mol\right)\)

\(m_{hh}=27a+56b=22\left(g\right)\left(1\right)\)

\(n_{H_2}=\dfrac{17.92}{22.4}=0.8\left(mol\right)\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(\Rightarrow1.5a+b=0.8\left(2\right)\)

\(\left(1\right),\left(2\right):a=0.4,b=0.2\)

\(m_{Al}=0.4\cdot27=10.8\left(g\right)\)

\(m_{Fe}=0.2\cdot56=11.2\left(g\right)\)

\(b.\)

\(n_{CuO}=\dfrac{68}{80}=0.85\left(mol\right)\)

\(n_{H_2}=n_{H_2O}=0.8\left(mol\right)\)

\(CuO+H_2\underrightarrow{^{t^0}}Cu+H_2O\)

\(BTKL:\)

\(m_{CuO}+m_{H_2}=m_{CR}+m_{H_2O}\)

\(\Rightarrow m_{CR}=68+0.8\cdot2-0.8\cdot18=55.2\left(g\right)\)

\(c.\)

\(n_{CuO\left(pư\right)}=x\left(mol\right)\)

\(CuO+H_2\underrightarrow{^{t^0}}Cu+H_2O\)

\(x...........x\)

\(m_{CR}=m_{CuO\left(dư\right)}+m_{Cu}=56-80x+64x=47.488\left(g\right)\)

\(\Rightarrow x=0.532\)

\(H\%=\dfrac{0.532\cdot80}{56}\cdot100\%=76\%\)