Đổi 448ml = 0,448l
Ta có:
\(n_{CO2}=\frac{0,448}{22,4}=0,02\left(mol\right)\)
\(n_{NaOH}=0,25.0,1=0,025\left(mol\right)\)
\(\frac{n_{NaOH}}{n_{CO2}}=\frac{0,025}{0,02}=1,25\)
\(2NaOH+CO_2\rightarrow Na_2CO_3+H_2O\left(1\right)\)
\(Na_2CO_3+CO_2+H_2O\rightarrow2NaHCO_3\left(2\right)\)
\(n_{Na2CO3\left(1\right)}=n_{CO2\left(1\right)}=\frac{n_{NaOH}}{2}=0,0125\left(mol\right)\)
\(n_{CO2\left(2\right)}=0,02-0,0125=0,0075\left(mol\right)\)
\(\Rightarrow n_{Na2CO3\left(2\right)}=0,0075\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{Na2CO3}=0,0125-0,0075=0,005\left(mol\right)\\n_{NaHCO3}=0,0075.2=0,015\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Na2CO3}=0,005.106=0,53\left(g\right)\\m_{NaHCO3}=0,015.84=1,26\left(g\right)\end{matrix}\right.\)
