|
|
$Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow$
a)$n_{H_2}=n_{Fe}=\frac{1,12}{56}=0,02 mol\Rightarrow V_{H_2}=22,4.0,02=0,448 lit$
b)$n_{H_2SO_4}=0,02 mol\Rightarrow m_{dd H_2SO_4 19,6\%}=\frac{0,02.98.100}{19,6}=10 gam$
c)$m_{FeSO_4}=152.0,02=3,04 gam$
$m_{dd}=1,12+10-2.0,02=11,08 gam\Rightarrow \%FeSO_4=27,44\%$
|
|
|
Trả lời 26-12-12 11:41 PM
viet130480
36K
1
5
13
414K
186K
401K
2
61% được chấp nhận
|
|