Bài 8:
PTHH: \(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)
Ta có: \(n_{H_2SO_4}=3n_{Al_2O_3}=3\cdot\dfrac{10,2}{102}=0,3\left(mol\right)\) \(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,3\cdot98}{9,8\%}=300\left(g\right)\)
Bài 9:
PTHH: \(SO_2+Ba\left(OH\right)_2\rightarrow BaSO_3\downarrow+H_2O\)
Đổi: 1 bar = 0,9869 atm
Ta có: \(n_{SO_2}=\dfrac{PV}{RT}=\dfrac{0,9869\cdot3,7185}{0,082\cdot\left(25+273\right)}\approx0,15\left(mol\right)=n_{Ba\left(OH\right)_2}\)
\(\Rightarrow V_{ddBa\left(OH\right)_2}=\dfrac{0,15}{0,5}=0,3\left(l\right)=300\left(ml\right)\)
