a, \(m_A=8+1,02=9,02\)
\(m_{giam}=m_O=9,02-7,74=1,28\left(g\right)\)
\(\rightarrow n_O=n_{H2\left(pu\right)}=0,08\left(mol\right)\)
\(n_{H2}=0,02\left(mol\right)\)
\(\rightarrow H=\frac{0,08.100}{0,2}=40\%\)
b, \(n_{CuO}=0,1\left(mol\right)\)
\(n_{F2eO3}=0,006\left(mol\right)\)
Gọi a là số mol của CuO (pu) , b là số mol Fe2O3 (pu)
\(\rightarrow\frac{a}{b}=\frac{0,1}{0,006}=\frac{50}{3}\Leftrightarrow3a-50b=0\left(1\right)\)
\(n_{H2\left(pu\right)}=n_{O\left(bi.khu\right)}=0,08\left(mol\right)\)
\(\rightarrow a+3b=0,08\)
Từ (1) và (2) \(\rightarrow\left\{{}\begin{matrix}a=0,068\\b=0,004\end{matrix}\right.\)
\(\rightarrow n_{Cu}=0,068,n_{Fe}=0,008\)
\(n_{CuO_{du}}=0,032\left(mol\right)\)
\(n_{Fe2O3_{du}}=0,002\left(mol\right)\)
\(\rightarrow m_{Cu}=4,352\left(g\right),m_{Fe}=0,448\left(g\right)\)
\(m_{Fe2O3_{du}}=0,32\left(g\right),m_{CuO_{du}}=2,56\left(g\right)\)