Answer:
The maximum amount of potassium carbonate that formed 49.68 g
Explanation:
According to question
2 KOH(aq) + CO₂(g) → K₂CO₃(aq) + H₂O(l)
16.2 grams of carbon dioxide are allowed to react with 45.0 grams of potassium hydroxide.
[tex]Moles(KOH)=\frac{Mass}{Molar mass}=\frac{45}{56}= 0.8 moles[/tex]
[tex]Moles(CO_{2} )=\frac{16.2}{44} = 0.36 moles[/tex]
[tex](KOH)\frac{Moles}{Stoichiometry}= \frac{0.8}{2} = 0.4\\(CO_{2} )\frac{Moles}{Stoichiometry}= \frac{0.36}{1} =0.36[/tex]
So, CO₂ is limiting reagent.
1 mole CO₂ produce 1 mole K₂CO₃
∴ 0.36 mole CO₂ produce 0.36 mole K₂CO₃ or (0.36 x 138)g = 49.68 g