What mass of KC_2H_3O_2KC2H3O2 is produced when 10.0 L CO_2CO2 is produced from potassium carbonate and excess acetic acid?

1 Answer
Alfredo T.
Feb 12, 2016

0,88 moles of KC_2H_3O_2KC2H3O2, that is 84 grams, are produced from 0.44 moles (10 L @ S.T.P.) of CO_2CO2.

Explanation:

The reaction equation is:

2CH_3COOH + K_2CO_3 -> 2KC_2H_3O_2 + CO_2 + H_2O2CH3COOH+K2CO3→2KC2H3O2+CO2+H2O

from which you can see that two units of potassium acetate (KC_2H_3O_2(KC2H3O2) are generated together with one mole of CO_2CO2.
Hence, provided one mol of CO_2CO2 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"10 L22.7 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 g98.15g/mol⋅0.88moles=86g.

Related questions
See all questions in Stoichiometry
Impact of this question
2726 views around the world
You can reuse this answer
Creative Commons License