Question

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

Answer 1

0,88 moles of `KC_2H_3O_2`, that is 84 grams, are produced from 0.44 moles (10 L @ S.T.P.) of `CO_2`.

Explanation:

The reaction equation is:

`2CH_3COOH + K_2CO_3 -> 2KC_2H_3O_2 + CO_2 + H_2O`

from which you can see that two units of potassium acetate `(KC_2H_3O_2`) are generated together with one mole of `CO_2`.
Hence, provided one mol of `CO_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: `"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 "g/mol" * 0.88 "moles" = 86 g`.