Question

Cho 10g CaCO3 phản ứng dd HCl 14,6%

a)Tính V_CO2

b)Tính m_ddHCl

c)Tính C% dd sau pứ

Answer 1

CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O

a) \(n_{CaCO_3}=\dfrac{10}{100}=0,1\left(mol\right)\)

Theo PT: \(n_{CO_2}=n_{CaCO_3}=0,1\left(mol\right)\)

\(\Rightarrow V_{CO_2}=0,1\times22,4=2,24\left(l\right)\)

\(m_{CO_2}=0,1\times44=4,4\left(g\right)\)

b) Theo PT: \(n_{HCl}=2n_{CaCO_3}=2\times0,1=0,2\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,2\times36,5=7,3\left(g\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{7,3\times100\%}{14,6\%}=50\left(g\right)\)

c) Theo PT: \(n_{CaCl_2}=n_{CaCO_3}=0,1\left(mol\right)\)

\(\Rightarrow m_{CaCl_2}=0,1\times111=11,1\left(g\right)\)

\(m_{dd}=10+50-4,4=55,6\left(g\right)\)

\(\Rightarrow C\%_{CaCl_2}=\dfrac{11,1}{55,6}\times100\%=19,96\%\)