3.0g of a mixture of potassium carbonate and potassium chloride were dissolved were dissolved in a 250cm\(^3\) standard flask. 25cm\(^3\) of this solution required 40.00cm\(^3\) of 0.1M HCl for neutralization. What is the percentage by weight of K\(_2\)CO\(_3\) in the mixture ? (K = 39, O = 16, C = 12)
60
72
82
89
92
Explanation
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Equation Of Reaction
2Hcl + K2CO3 ----> 2Kcl + CO2 + H2O
Molar Mass Of Base, K2CO3 =138g/mol
CaVa/ CbVb=na/nb
0.1x 40/ Cbx25 = 2/1
50cb =4
Cb= 4/50
Cb=0.08M
Therefore 0.08x 138
=11.8g
Mass Is 3g In 25cm3 Of The Mixture
3/25x100
=12
Percentage Of K2CO3
11.04/12x 100
=92%
K2CO3 + 2HCl ---------> 2Kcl + H2O + CO2
Using
CaVa/CbVb = na/nb
0.1× 40/25×Cb = 2/1
Cb = 0.08mol/dm3
Therefore :
0.08mol/dm3 diss in 1000cm3 of mixture
X mol will dissolve in 250cm3
Cross multipling we have that
X = 0.02mol
Using :
n =m/M
Therefore:
m = n×M
m = 0.02mol× 138g/mol( molar mass of K2CO3)
m = 2.79g of K2CO3
% by mass = 2.79g/3g ×100
= 92%
I hope it's short and simple.
God of Mercy. I wasted so much time on this question not knowing it was not correct
The real question 👇🏾👇🏾👇🏾
there's a mistake in the question it say potassium carbonate and potassium chloride instead of hydrogen chloride
