PT: \(Fe_2O_3+6HNO_3\rightarrow2Fe\left(NO_3\right)_3+3H_2O\)
\(CuO+2HNO_3\rightarrow Cu\left(NO_3\right)_2+H_2O\)
Giả sử: \(\left\{{}\begin{matrix}n_{Fe_2O_3}=x\left(mol\right)\\n_{CuO}=y\left(mol\right)\end{matrix}\right.\)
⇒ 160x + 80y = 24 (1)
Có: \(m_{HNO_3}=800.6,3\%=50,4\left(g\right)\Rightarrow n_{HNO_3}=\dfrac{50,4}{63}=0,8\left(mol\right)\)
Theo PT: \(n_{HNO_3}=6n_{Fe_2O_3}+2n_{CuO}=6x+2y\left(mol\right)\)
⇒ 6x + 2y = 0,8 (2)
Từ (1) và (2) ⇒ x = y = 0,1 (mol)
a, mFe2O3 = 0,1.160 = 16 (g)
mCuO = 8 (g)
b, Có: m dd sau pư = mX + m dd HNO3 = 24 + 800 = 824 (g)
Theo PT: \(\left\{{}\begin{matrix}n_{Fe\left(NO_3\right)_3}=2n_{Fe_2O_3}=0,2\left(mol\right)\\n_{Cu\left(NO_3\right)_3}=n_{CuO}=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{Fe\left(NO_3\right)_2}=\dfrac{0,2.242}{824}.100\%\approx5,87\%\\C\%_{Cu\left(NO_3\right)_2}=\dfrac{0,1.188}{824}.100\%\approx2,28\%\end{matrix}\right.\)
Bạn tham khảo nhé!
\(n_{Fe_2O_3}=a\left(mol\right),n_{CuO}=b\left(mol\right)\)
\(m_X=160a+80b=24\left(g\right)\left(1\right)\)
\(n_{HNO_3}=\dfrac{800\cdot6.3\%}{63}=0.8\left(mol\right)\)
\(Fe_2O_3+6HNO_3\rightarrow2Fe\left(NO_3\right)_3+6H_2O\)
\(CuO+2HNO_3\rightarrow Cu\left(NO_3\right)_2+H_2O\)
\(n_{HNO_3}=6a+2b=0.8\left(mol\right)\left(2\right)\)
\(\left(1\right),\left(2\right):a=0.1,b=0.1\)
\(m_{Fe_2O_3}=0.1\cdot160=16\left(g\right),m_{CuO}=8\left(g\right)\)
\(m_{dd}=24+800=824\left(g\right)\)
\(C\%_{Fe\left(NO_3\right)_3}=\dfrac{0.2\cdot242}{824}\cdot100\%=5.87\%\)
\(C\%_{Cu\left(NO_3\right)_2}=\dfrac{0.1\cdot188}{824}\cdot100\%=2.3\%\)
