[Hoá học 9] Tính C%

T

thupham22011998

Ta có: $n K_2CO_3=4,74:158=0,03 mol$

$K_2CO_3+2HCl--->2KCl+CO_2+H_2O$
0,03..............0,06

-->C%$HCl=1,46$%
 
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H

hocgioi2013

$K_2SO_3+2HCl-->2KCl+H_2O+SO_2$
chuyển về mol $K_2SO_3$ rồi kéo qua $HCl$
tính C% bình thường
 
K

kimphuc2112

nK_2SO_4=\frac{4,74}{158}=0,03(mol)
pt:K^2SO_3+2HCl\Rightarrow2KCll+SO_2+H_2O
0,03mol 0,06mol
mHCl=0,06.33=1,98(g)
C%HCl=\frac{mHCl}{m_ddHCl}.100%
={1,98}{150}.100%
=1,32(%)
 
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