Hóa 10

V

vuphong0707

T

thupham22011998

a,Ta có: $n CO_2=0,2 mol -->n CaCO_3=0,2 mol$

$-->m CaCO_3=20g -->m CaO=16,8g$

-->%$mCaCO_3$=54,35%

-->%$m CaO$=45,65%

b, Ta có: $n CaO=0,3 mol$

Theo các pthh, ta có: $n HCl=1 mol ; n CaCl_2=0,5 mol$

$-->C_M CaCl_2=0,1M$

$-->C_M HCl=0,2M$
 
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