Đáp án:
$a,V_{CO_2}=4,48l.$
$b,m_{ddCH_3COOH}=160g.$
$c,m_{(CH_3COO)_2Ca}=31,6g.$
Giải thích các bước giải:
$a,PTPƯ:CaCO_3+2CH_3COOH\xrightarrow{} (CH_3COO)_2Ca+H_2O+CO_2$
$n_{CaCO_3}=\dfrac{20}{100}=0,2mol.$
$Theo$ $pt:$ $n_{CO_2}=n_{CaCO_3}=0,2mol.$
$⇒V_{CO_2}=0,2.22,4=4,48l.$
$b,Theo$ $pt:$ $n_{CH_3COOH}=2n_{CaCO_3}=0,4mol.$
$⇒m_{CH_3COOH}=0,4.60=24g.$
$⇒m_{ddCH_3COOH}=\dfrac{24}{15\%}=160g.$
$c,Theo$ $pt:$ $n_{(CH_3COO)_2Ca}=n_{CaCO_3}=0,2mol.$
$⇒m_{(CH_3COO)_2Ca}=0,2.158=31,6g.$
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