Cần gấp

P

pinkylun

$2Al+3H_2SO_4---->Al_2(SO_4)_3+3H_2$

a) $n_{H_2}=\dfrac{3,36}{22,4}=0,15 mol$

$=>n_{Al}=0,1mol$

$=>m_{Al}=0,1.27=2,7g$

b) $ n_{H_2SO_4}=0,15mol$

$=>V_{ d d}=\dfrac{n}{C_M}=\dfrac{0,15}{1}=0,15l=150ml$

 
C

chaugiang81

giải luôn câu c nhé :)
$nAl_2(SO_4)_3$ la : 0.15 *1 /3=0.05 mol.
$CM_{Al_2(SO4)_3}$ là:
CM= $\dfrac{0.05}{0.15}$= 0.33M

Chất rắn ko có $C_M$
 
Last edited by a moderator:
You must log in or register to reply here.
Top Bottom