a)Theo bài ta có;
\(n_{Fe_2O_3}=\dfrac{m_{Fe_2O_3}}{M_{Fe_2O_3}}=\dfrac{3,2}{160}=0,02\left(mol\right)\)
pthh:\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
số mol__0,2_____1,2
Theo bài và pthh ta có:
\(n_{HCl}=6\cdot n_{Fe_2o_3}=6\cdot0,2=1,2\left(mol\right)\)
\(\Rightarrow m_{HCl}=n_{HCl}\cdot M_{HCl}=1,2\cdot36,5=43,8\left(g\right)\)
\(\Rightarrow C\%=\dfrac{m_{HCl}}{m_{ddHCl}}=\dfrac{43,8}{200}=21,9\%\)
b)
\(m_{HCl}=\dfrac{C\%\cdot m_{ddHCl2,5\%}}{100\%}=\dfrac{2,5\%\cdot200}{100\%}=5\left(g\right)\)
\(\Rightarrow n_{HCl}=\dfrac{m_{HCl}}{M_{HCl}}=\dfrac{5}{36,5}\approx0,137\left(mol\right)\)
Ta có :
\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
số mol bđ: 0,2______0,137
số mol tgpư: 0,022_____0,137
số mol dư: 0,178______0
⇒ Sau phản ứng Fe2O3 còn dư,HCl tgpuw hết.
⇒DD sau pư gồm FeCl3 và Fe2O3 dư
Ta có
\(n_{FeCl_3}=\dfrac{1}{3}\cdot n_{HCl}=\dfrac{1}{3}\cdot0,137\approx0.045\left(mol\right)\)
\(\Rightarrow m_{FeCl_3}=n_{FeCl_3}\cdot M_{FeCl_3}=0,045\cdot162,5=7,3125\left(g\right)\)
\(m_{Fe_2O_3dư}=n_{Fe_2O_3dư}\cdot M_{Fe_2O_3}=0,178\cdot160=28,48\left(g\right)\)
\(m_{ddspuw}=m_{Fe_2O_3}+m_{ddHCl}=3,2+200=203,2\left(g\right)\)
\(\Rightarrow C\%_{FeCl_3}=\dfrac{m_{FeCl3}\cdot100\%}{m_{ddspuw}}=\dfrac{7,3125\cdot100\%}{203,2}\approx3,6\%\)
\(\Rightarrow C\%_{Fe2O3dư}=\dfrac{m_{Fe2O3duw}\cdot100\%}{m_{ddspuw}}=\dfrac{28,48\cdot100\%}{203,3}=14,01\%\)
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