a)
$CaCO_3 + 2HCl \to CaCl_2 + CO_2 + H_2O$
b)
$n_{CO_2} = n_{CaCO_3} = \dfrac{20}{100} = 0,2(mol)$
$V_{CO_2} = 0,2.22,4 = 4,48(lít)$
c)
Sau phản ứng :
$m_{dd} = 20 + 200 - 0,2.44 = 211,2(gam)$
$n_{CaCl_2} = n_{CaCO_3} = 0,2(mol)$
$C\%_{CaCl_2} = \dfrac{0,2.111}{211,2}.100\% = 10,51\%$
a, PTHH: \(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)
b, \(n_{CO_2}=n_{CaCl_2}=n_{CaCO_3}=\dfrac{20}{100}=0,2\left(mol\text{}\right)\)
\(\Rightarrow V_{CO_2}=0,24,48\left(l\right)\)
c, \(m_{ddsaupu}=200+20-44.0,2=211,2\left(g\right)\)
\(\Rightarrow C\%=\dfrac{0,2.111}{211,2}.100\%=10,51\%\)
\(n_{CaCO_3}=\dfrac{20}{100}=0,2\left(mol\right)\\ a.CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\\ b.n_{CO_2}=n_{CaCl_2}=n_{CaCO_3}=0,2\left(mol\right)\\ m_{ddCaCl_2}=20+200-0,2.44=211,2\left(g\right)\\ V_{CO_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\\ c.m_{CaCl_2}=0,2.111=22,2\left(g\right)\\ C\%_{ddCaCl_2}=\dfrac{22,2}{211,2}.100\approx10,511\%\)
\(n_{CaCO3}=\dfrac{20}{100}=0,2\left(mol\right)\)
a) Pt : \(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O|\)
1 2 1 1 1
0,2 0,2 0,2
b) \(n_{CO2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
\(V_{CO2\left(dktc\right)}=0,2.22,4=4,48\left(l\right)\)
c) \(n_{CaCl2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
⇒ \(m_{CaCl2}=0,2.111=22,2\left(g\right)\)
Sau phản ứng :
\(m_{dd}=20+200-\left(0,2.44\right)\)
= 211,2 (g)
\(C_{CaCl2}=\dfrac{22,2.100}{211,2}=10,51\)0/0
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