a, \(CaCO_3+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Ca+CO_2+H_2O\)
b, \(n_{CaCO_3}=\dfrac{10}{100}=0,1\left(mol\right)\)
Theo PT: \(n_{CO_2}=n_{CaCO_3}=0,1\left(mol\right)\Rightarrow V_{CO_2}=0,1.22,4=2,24\left(l\right)\)
c, \(n_{\left(CH_3COO\right)_2Ca}=n_{CaCO_3}=0,1\left(mol\right)\)
\(n_{CH_3COOH}=2n_{CaCO_3}=0,2\left(mol\right)\Rightarrow m_{CH_3COOH}=0,2.60=12\left(g\right)\)
\(\Rightarrow m_{ddCH_3COOH}=\dfrac{12}{5\%}=240\left(g\right)\)
Ta có: m dd sau pư = 10 + 240 - 0,1.44 = 245,6 (g)
\(\Rightarrow C\%_{\left(CH_3COO\right)_2Ca}=\dfrac{0,1.158}{245,6}.100\%\approx6,43\%\)
a)
$CaCO_3 + 2CH_3COOH \to (CH_3COO)_2Ca + CO_2 + H_2O$
b)
$n_{CaCO_3} = \dfrac{10}{100} = 0,1(mol)$
Theo PTHH :
$n_{CO_2} = n_{CaCO_3} = 0,1(mol)$
$V_{CO_2} = 0,1.22,4 = 2,24(lít)$
c)
$n_{CH_3COOH} = 2n_{CaCO_3} = 0,2(mol)$
$m_{dd\ CH_3COOH} = \dfrac{0,2.60}{5\%} = 240(gam)$
$m_{dd\ sau\ pư} = 10 + 240 - 0,1.44 = 245,6(gam)$
$n_{(CH_3COO)_2Ca} = n_{CaCO_3} = 0,1(mol)$
$C\%_{(CH_3COO)_2Ca} = \dfrac{0,1.158}{245,6}.100\% = 6,4\%$
