mdd=5,85+8+200=213,85g
\(C\%_{NaCl}=\dfrac{5,85.100}{213,85}\approx2,74\%\)
\(C\%_{NaOH}=\dfrac{8.100}{213,85}\approx3,74\%\)
\(n_{CO_2}=\dfrac{4,48}{22,4}=0,2mol\)
\(m_{CO_2}=0,2.44=0,88g\)
\(n_{CO_2}=\dfrac{3,36}{22,4}=0,15mol\)
\(n_{NaOH}=\dfrac{50.12}{40.100}=0,15mol\)
\(n_{CO_2}=n_{NaOH}\)\(\rightarrow\)Phản ứng vừa đủ sinh ra NaHCO3
CO2+NaOH\(\rightarrow\)NaHCO3
\(n_{NaHCO_3}=0,15mol\rightarrow m_{NaHCO_3}=0,15.84=12,6g\)
- Số phân tử CO2=0,2.N=0,2.6,02.1023=1,204.1023 phân tử