nCaCO3 = \(\dfrac{m}{M}\)=\(\dfrac{10}{100}\)= 0,1 (mol)
a. PTHH:
CaCO3 + 2HCl → CaCl2 + CO2 + H2O
1 : 2 : 1 : 1 : 1 (mol)
0,1 : 0,2 : 0,1 : 0,1 : 0,1 (mol)
b. mHCl = n.M = 0,2.36,5 = 7,3 (g)
mdd HCl = \(\dfrac{m_{ct}.100\%}{C\%}\)=\(\dfrac{7,3.100}{10}\)= 73 (g)
mdd (pư) = 10 + 73 = 83 (g)
c. VCO2 (đktc) = n.22,4 = 0,1.22,4 = 2,24 (l)
d. mCaCl2 = n.M = 0,1.111 = 11,1 (g)
mdd (sau pư) = mdd (pư) - mCO2 = 83 + (0,1.44) = 87,4 (g)
C% = \(\dfrac{m_{ct}}{m_{dd}}\).100% = \(\dfrac{11,1}{87,4}\).100% = 12,7%
a) PTHH: CaCO3 + 2HCl → CaCl2 + CO2↑ + H2O
b) \(n_{CaCO_3}=\dfrac{10}{100}=0,1\left(mol\right)\)
Theo PT: \(n_{HCl}=2n_{CaCO_3}=2\times0,1=0,2\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,2\times36,5=7,3\left(g\right)\)
\(\Rightarrow m_{ddHCl}=\dfrac{7,3\times100\%}{10\%}=73\left(g\right)\)
c) Theo PT: \(n_{CO_2}=n_{CaCO_3}=0,1\left(mol\right)\)
\(\Rightarrow V_{CO_2}=0,1\times22,4=2,24\left(l\right)\)
\(m_{CO_2}=0,1\times44=4,4\left(g\right)\)
d) Theo PT: \(n_{CaCl_2}=n_{CaCO_3}=0,1\left(mol\right)\)
\(\Rightarrow m_{CaCl_2}=0,1\times111=11,1\left(g\right)\)
\(m_{dd}=10+73-4,4=78,6\left(g\right)\)
\(\Rightarrow C\%_{CaCl_2}=\dfrac{11,1}{78,6}\times100\%=14,12\%\)
