Bài 1:
PTHH: \(SO_3+Ba\left(OH\right)_2\rightarrow BaSO_4\downarrow+H_2O\)
Ta có: \(n_{SO_3}=\dfrac{6,72}{22,4}=0,3\left(mol\right)=n_{BaSO_4}=n_{Ba\left(OH\right)_2}\)
\(\Rightarrow\left\{{}\begin{matrix}m_{ddBa\left(OH\right)_2}=\dfrac{0,3\cdot171}{17,1\%}=300\left(g\right)\\m_{BaSO_4}=0,3\cdot233=69,9\left(g\right)\end{matrix}\right.\)
Bài 2:
PTHH: \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
Ta có: \(n_{Fe_2O_3}=\dfrac{3,2}{160}=0,02\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,12\left(mol\right)\\n_{FeCl_3}=0,04\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}C_{M_{HCl}}=\dfrac{0,12}{0,2}=0,6\left(M\right)\\m_{FeCl_3}=0,04\cdot162,5=6,5\left(g\right)\end{matrix}\right.\)
Bài 5:
PTHH: \(ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\)
Ta có: \(\left\{{}\begin{matrix}n_{ZnO}=\dfrac{8,1}{81}=0,1\left(mol\right)\\n_{H_2SO_4}=\dfrac{100\cdot14,7\%}{98}=0,15\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) Axit còn dư, tính theo ZnO
\(\Rightarrow\left\{{}\begin{matrix}n_{ZnSO_4}=0,1\left(mol\right)\\n_{H_2SO_4}=0,05\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}C\%_{ZnSO_4}=\dfrac{0,1\cdot161}{8,1+100}\cdot100\%\approx14,9\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{0,05\cdot98}{8,1+100}\cdot100\%\approx4,53\%\end{matrix}\right.\)
Bài 3:
PTHH: \(Na_2O+H_2O\rightarrow2NaOH\)
Theo PTHH: \(n_{NaOH}=2n_{Na_2O}=2\cdot\dfrac{6,2}{62}=0,2\left(mol\right)\)
\(\Rightarrow C\%_{NaOH}=\dfrac{0,2\cdot40}{200}\cdot100\%=4\%\)
Bài 4:
PTHH: \(P_2O_5+3H_2O\rightarrow2H_3PO_4\)
Ta có: \(n_{H_3PO_4}=2n_{P_2O_5}=2\cdot\dfrac{14,2}{142}=0,2\left(mol\right)\) \(\Rightarrow C_{M_{H_3PO_4}}=\dfrac{0,2}{0,2}=1\left(M\right)\)
