Al2O3 + 3H2SO4 -> Al2(SO4)3 + 3H2O (1)
nAl2O3=0,1(mol)
Từ 1:
nAl2(SO4)3=nAl2O3=0,1(mol)
mAl2(SO4)3=342.0,1=34,2(g)
C% dd Al2(SO4)3=\(\dfrac{34,2}{300+10,2}.100\%=11\%\)
Al2O3 + 3H2SO4 ➜ Al2(SO4)3 + 3H2O
\(n_{Al_2O_3}=\dfrac{10,2}{102}=0,1\left(mol\right)\)
Theo PT: \(n_{Al_2\left(SO_4\right)_3}=n_{Al_2O_3}=0,1\left(mol\right)\)
\(\Rightarrow m_{Al_2\left(SO_4\right)_3}=0,1\times342=34,2\left(g\right)\)
\(\Sigma m_{dd}=10,2+300=310,2\left(g\right)\)
\(\Rightarrow C\%_{ddAl_2\left(SO_4\right)_3}=\dfrac{34,2}{310,2}\times100\%\approx11\%\)
Al2O3 + 3H2SO4 ➜ Al2(SO4)3 + 3H2O
\(n_{Al_2O_3}=\dfrac{10,2}{102}=0,1\left(mol\right)\)
\(\Sigma m_{dd}=10,2+300=310,2\left(g\right)\)
Theo PT: \(n_{Al_2\left(SO_4\right)_3}=n_{Al_2O_3}=0,1\left(mol\right)\)
\(\Rightarrow m_{Al_2\left(SO_4\right)_3}=0,1\times342=34,2\left(g\right)\)
\(\Rightarrow C\%_{ddAl_2\left(SO_4\right)_3}=\dfrac{34,2}{310,2}\times100\%\approx11\%\)