# What mass of KC_2H_3O_2 is produced when 10.0 L CO_2 is produced from potassium carbonate and excess acetic acid?

Feb 12, 2016

0,88 moles of $K {C}_{2} {H}_{3} {O}_{2}$, that is 84 grams, are produced from 0.44 moles (10 L @ S.T.P.) of $C {O}_{2}$.

#### Explanation:

The reaction equation is:

$2 C {H}_{3} C O O H + {K}_{2} C {O}_{3} \to 2 K {C}_{2} {H}_{3} {O}_{2} + C {O}_{2} + {H}_{2} O$

from which you can see that two units of potassium acetate (KC_2H_3O_2) are generated together with one mole of $C {O}_{2}$.
Hence, provided one mol of $C {O}_{2}$ at S.T.P. (standard temperature and pressure: T = 0 °C, P = 1 bar) have an approximate volume of 22.7 liters, the volume of 10 L would correspond to: $\text{10 L"/"22.7 L/mol" = "0.44 moles}$.

Then the chemical amount of potassium acetate are 0.88 mol.
The molar mass of potassium acetate is 98.15 g/mol.

So, the mass of potassium acetate is $98.15 \text{g/mol" * 0.88 "moles} = 86 g$.