Answer:
a) mass of the H2CO3 produced:
given:
Mass of sample = 3.24 g
Mass of Na2CO3 obtained after decomposition = 2.19 g
Solution :
Molar mass of NaHCO3 = 84
reaction:
2NaHCO3 → Na2CO3 + H2CO3
so it is clear that 2 mole of NaHCO3 gives 1 mole of Na2CO3 and H2CO3
Now, ICE table for the reaction is :
NaHCO3 Na2CO3 H2CO3
I 3.24/84 0 0
C -2x +x +x
E 3.24/84 -2x x x
As NaHCO3 is completely decomposed so final Concentration of NaHCO3 is zero.
=> 3.24/84 -2x = 0
=> 2x = 3.24/84
=> x = 1.62/84
The new ICE table is :
NaHCO3 Na2CO3 H2CO3
I 3.24/84 0 0
C -2x = -2(1.62/84) +x = 1.62/84 +x = 1.62/84
E 0 1.62/84 1.62/84
From the above ICE table,
it is found that (1.62/84 ) moles of H2CO3 is obtained.
Since,
The molar mass of H2CO3 is 62
=> Mass of H2CO3 obtained = moles × molar mass
=> Mass of H2CO3 obtained = (1.62 /84 ) × 62
= 1.19 grams
Mass of H2CO3 experimentally :
Mass of reactants = mass of products
=> Mass of sample = mass of Na2CO3 + mass of H2CO3
=> Mass of H2CO3 = mass of sample - mass of Na2CO3
= 3.24 - 2.19 = 1.05 g
b) Experimental mass = 1.05 g
Theoretical mass = 1.19 g
Percentage yield of H2CO3 = Experimental mass × 100 / Theoretical mass
= 1.05 × 100 /1.19
= 88.23 %
