Bài 7:
a) \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
b) \(n_{Al}==\dfrac{4,05}{27}=0,15\left(mol\right)\); \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
Theo PTHH: \(n_{Al\left(pư\right)}=\dfrac{2}{3}.n_{H_2}=0,1\left(mol\right)\)
=> mAl(pư) = 0,1.27 = 2,7 (g)
c) Theo PTHH: \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{3}.n_{H_2}=0,05\left(mol\right)\)
=> \(m_{Al_2\left(SO_4\right)_3}=0,05.342=17,1\left(g\right)\)
mAl(dư) = 4,05 - 2,7 = 1,35 (g)
d) Theo PTHH: \(n_{H_2SO_4}=n_{H_2}=0,15\left(mol\right)\)
=> \(m_{H_2SO_4}=0,15.98=14,7\left(g\right)\)
Bài 10:
a) \(n_{Fe_2O_3}=\dfrac{24}{160}=0,15\left(mol\right)\); \(n_{H_2SO_4}=2,5.0,2=0,5\left(mol\right)\)
b)
PTHH: \(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)
Xét tỉ lệ: \(\dfrac{0,15}{1}< \dfrac{0,5}{3}\) => Fe2O3 hết, H2SO4 dư
PTHH: \(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)
0,15----->0,45--------->0,15
=> \(n_{H_2SO_4\left(dư\right)}=0,5-0,45=0,05\left(mol\right)\)
c) \(\left\{{}\begin{matrix}C_{M\left(Fe_2\left(SO_4\right)_3\right)}=\dfrac{0,15}{0,2}=0,75M\\C_{M\left(H_2SO_4.dư\right)}=\dfrac{0,05}{0,2}=0,25M\end{matrix}\right.\)
