a) \(n_{H_2}=\frac{7,84}{22,4}=0,35\left(mol\right)\)
\(m_{H_2}=0,35.18=6,3\left(g\right)\)
b) Gọi x là số mol của \(CuO\), y là số mol của \(Fe_2O_3\)
Theo đề bài, ta có: \(80x+160y=20\left(1\right)\)
\(PTHH\left(1\right):CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
\(\left(mol\right)\)________x________________x
\(PTHH\left(2\right):Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
\(\left(mol\right)\)______y_________3y___2y_____3y
\(TừPT\left(1\right)và\left(2\right)\Rightarrow x+3y=\frac{7,84}{22,4}=0,35 \left(2\right)\)
\(Từ\left(1\right)và\left(2\right)tacóhpt:\left\{{}\begin{matrix}80x+160y=20\\x+3y=0,35\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0,05\left(mol\right)\\y=0,1\left(mol\right)\end{matrix}\right.\)
\(m_{hhkl}=m_{Cu}+m_{Fe}=64x+56.2y=14,4\left(g\right)\)
c) \(PTHH\left(3\right):Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(\left(mol\right)\)________0,35___0,7______0,35___0,35
\(m_{Zn}=0,35.65=22,75\left(g\right)\)
\(m_{HCl}=0,7.36,5=25,55\left(g\right)\)
\(m_{ddHCl}=\frac{25,55.100\%}{20\%}=127,75\left(g\right)\)
