Question

hòa tan 10,2g Al2O3 vào 200g dd H2SO4 19,6%

a) Viết PTHH

b) Tính C% dd sau PƯ

*đề ko cho C% của Al2O3

Answer 1

a) \(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)

b) \(n_{Al_2O_3}=\dfrac{10,2}{102}=0,1\left(mol\right)\); \(n_{H_2SO_4}=\dfrac{200.19,6\%}{98}=0,4\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,4}{3}\) => Al₂O₃ hết, H₂SO₄ dư

PTHH: \(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)

               0,1------->0,3--------->0,1

\(\left\{{}\begin{matrix}m_{Al_2\left(SO_4\right)_3}=0,1.342=34,2\left(g\right)\\m_{H_2SO_4\left(dư\right)}=\left(0,4-0,3\right).98=9,8\left(g\right)\end{matrix}\right.\)

m_{dd sau pư} = 10,2 + 200 = 210,2 (g) 

\(\left\{{}\begin{matrix}C\%_{Al_2\left(SO_4\right)_3}=\dfrac{34,2}{210,2}.100\%=16,27\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{9,8}{210,2}.100\%=4,66\%\end{matrix}\right.\)