At 200 °C, potassium hydrogencarbonate decomposes according to the following equation.
2KHCO3 → K2CO3 +H2O+CO2
What is the loss in mass when 50.0 g of potassium hydrogencarbonate are heated at 200°C to constant mass?
(A,:
H = 1; C = 12;
0 = 16; K = 39)
A
11.0g
B
15.5g
C
22.0 g
D
31.0g
Hi!
In this particular chemical reaction the only non-gaseous product is K2CO3 as at 200 degrees celcius water is water vapour and CO2 is also gas. Hence, the loss in mass will be the difference b/w our starting mass of KHCO3 and our ending mass of K2CO3.
n(moles) KHCO3 = 50/100=0.5 moles. This means that the moles of K2CO3 is half the moles KHCO3 as they are in 2:1 ratio. so, moles K2CO3=0.25 moles.
Mr K2CO3 = 0.25 x 138=34.5g.
so, the loss in mass is 50-34.5=15.5g.
Answer is thus B.
Hope it helps:)
thanks so much, i forgot about CO2 and H2O being gas ahahhah.
