测试

链剑心

\[\iiint\]

链剑心

\[\mathrm{NaHCO_{3}}+\mathrm{CH_3COOH}\longrightarrow\mathrm{CH_3COONa}+\mathrm{H_2O}+\mathrm{CO_2} \]

链剑心

\[\mathrm{NaHCO_{3}}+\mathrm{CH_3COOH}\longrightarrow\mathrm{CH_3COONa}+\mathrm{H_2O}+\mathrm{CO_2}\uparrow \]

链剑心

\[\mathrm{Molar\,mass\,of\,NaHCO_3}=(24.31+1.008+12.01+3\times16.00)\mathrm{g/mol}=85.328\mathrm{g/mol}\]

\[\text{Molar mass of acetic acid}=(12.01+3\times1.008+12.01+16.00+16.00+1.008)\mathrm{g/mol}=60.052\mathrm{g/mol}\]

\[\text{Molar mass of carbon dioxide}=(12.01+32.00)\mathrm{g/mol}=44.01\mathrm{g/mol}\]

\[\text{Molar mass of water}=(1.008\times2+16.00)\mathrm{g/mol}=18.016\mathrm{g/mol}\]

链剑心

How many grams of $\mathrm{CO_2}$ can be produce, if $x\,\rm{ml}$ of acetic acid involves reacting?

solution:
The mass of sodium bicarbonate
\[m=\rho V=2.20x\,\rm g\]
The mass of carbon dioxide
\[\frac{2.20x\,\rm g}{85.328\mathrm{g/mol}}\times\frac{1}{1}\times44.01\mathrm{g/mol}=1.1347x\,\rm g\]

Inversely, the mass of Sodium bicarbonate can be calculated by

\[\mathrm{mass\,CO_2\times\frac{22000}{11347}}=\mathrm{mass\,NaHCO_3}\]