hoá

K

kagaminex

$C_2H_4+3O_2--t^o-->2CO_2+2H_2O$

$Ca(OH)_2+CO_2---->CaCO_3+H_2O$

$n_{Ca(OH)_2}=0,15<n_{CO_2}=0,2$

$=>m_{CaCO_3}=0,15*100=15(g)$

$m_{CO_2}+m_{H_2O}=12,4(g)$

=>khối lượng dung dung giảm và giảm 15-12,4=2,6(g)

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