Question

A 3.24-gram sample of NaHCO3 was completely decomposed in an experiment. 2NaHCO3 → Na2CO3 + H2CO3 In this experiment, carbon dioxide and water vapors combine to form H2CO3. After decomposition, the Na2CO3 had a mass of 2.19 grams. Determine the mass of the H2CO3 produced. Calculate the percentage yield of H2CO3 for the reaction. Show your work or describe the calculation process in detail.

Answer 1

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 %