K2CO3(s) + 2HCl(aq) → 2KCl(aq) + H2O(l) + CO2(g)?
Please Help!!!!
Excess hydrochloric acid is reacted with 4.12 g potassium carbonate. If the reaction occurs at STP, what volume of carbon dioxide gas is produced?
K2CO3(s) + 2HCl(aq) → 2KCl(aq) + H2O(l) + CO2(g)
More Questions From This User See All
Answers & Comments
Verified answer
moles K2CO3 = mass / molar mass
= 4.12 g / 138.2 g/mol
= 0.02981 mol
The balanced eqaution tells you that 1 mole K2CO3 forms 1 mole CO2
So moles CO2 = moles K2CO3 = 0.02981 mol
At STP 1 mole of any ideal gas occupies a volume of 22.4 L
So volume CO2 at STP = 22.4 L/mol x 0.02981 mol
= 0.668 L
= 668 ml