BMAT 2006 Section 2 Question 13

At 200 °C, potassium hydrogencarbonate decomposes according to the following equation. What is the loss in mass when 50.0 g of potassium hydrogencarbonate are heated at 200 °C to constant mass?

Answer:

Mr of 2KHCO₃ = 2(39 + 1 + 12 + 48) = 200
moles of KHCO₃ = m/Mr = 50 ÷ 200 = 0.25

moles of K₂CO₃ = 0.25 ÷ 2 = 0.125
mass of K₂CO₃ = 0.125 x 138 = 34.5g

The loss in mass is calculated as the mass of the gases (H₂O and CO₂) produced.
Loss in mass = Mass of KHCO₃ – Mass of K₂CO₃ = 50-34.5 = 15.5g

Sami Qamar

I’m Sami Qamar. I’m a YouTuber, Blogger, and first year med student.

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