2Al +3H2SO4 ->Al2[SO4]3 +3H2
ta có nH2= 3,36:22,4=0,15 mol
theo pthh;nAl=2/3 nH2=0.1 mol
=>mAl=0,1.27=2.7g
theo pthh;nH2SO4 =nH2=0,15 mol
=>mH2SO4 =0,15 .98=14,7 g
=>mddH2SO4 =14,7.100:49=30 g
mH2=0,15.2=0.3g
=>mdd sau pu =2,7+30-0,3=32,4g
theo pthh nAl2[SO4]3 =1/3 nH2 =0,05 mol
=>mAl2[SO4]3 =0,05.342=17,1 g
=>C% DD sau pu =17,1:32,4.100=52,78%
ta co pthh:2Al+3H\(_2\)SO\(_4\)\(\rightarrow\)Al\(_2\)(SO\(_4\))+3H\(_2\)
ta có n\(_{H_2}\)=3,36\(\div22,4\)
=0,15(mol)
theo pthh ta có :n\(_{Al}\)=\(\dfrac{2}{3}\)n\(_{H_2}\)
=\(\dfrac{2}{3}\times0,15\)
=0,1(mol)
ta lại có :n\(_{H_2}\)\(_{SO_4}\)=n\(_{H_2}\)=0,15(mol)
\(\Rightarrow\)m\(_{H_2}\)\(_{SO_4}\)=0,15\(\times\)98
=14,7
\(\Rightarrow\)mdd=(14,7\(\times\)100)\(\div\)49%
=30(g)
ta có n\(_{Al_2\left(SO_4\right)_3}\)=\(\dfrac{1}{3}\)n\(_{H_2}\)
=\(\dfrac{1}{3}\)\(\times\)0,15
=0,05(mol)
\(\Rightarrow\)m=0,05*342
=17,1(g)
\(\Rightarrow\)C%=\(\dfrac{17,1}{32,4}\)\(\times\)100
=52,7(%)
n\(H_2\)=\(\dfrac{3,36}{22,4}\)=0,15(mol)
2Al + 3H2SO4 -> Al2(SO4)3 + 3H2
| 2 | 3 | 1 | 1 |
| 0,1 | 0,15 | 0,05 | 0,15 |
mH2SO4 = 0,15.98=14,7(g)
mddH2SO4 = 14,7:49%=30(g)
mAl2(SO4)3 = 0,05.342=17,1(g)
mAl= 0,1.27=2,7(g)
mddAl2(SO4)3= 30+2,7-3,36=29,34(g)
C%= \(\dfrac{17,1}{29,34}\).100%=58,3%