The statement of "the mass of the H2CO3 produced is 1.31 g and the the percentage yield of H2CO3 is 96.43%."
What is percentage yield?
The percentage yield is the percent ratio of actual yield to the theoretical yield. It is obtained by dividing the observed yield by the theoretical yield and multiplying by 100%.
Given:
2NaHCO3 → Na2CO3 + H2CO3
3.81 g of NaHCO3, 2.86 g of Na2CO3
Solution:
One mole of sodium carbonate and one mole of hydrogen carbonate are produced from two moles of sodium bicarbonate.
Gram to mole conversion,
[tex]3.81\;.\frac{1\;mole}{84\;g}\;=\;0.0453\;mol\;of\;NaHCO3\\2.86\;.\frac{1\;mole}{105.98\;g}\;=\;0.0269\;mol\;of\;Na2CO3[/tex]
Initially to find the mass of H2CO3, moles of H2CO3 can be calculated by using the determined moles of NaHCO3.
[tex]0.0453\;mol\;NaHCO3.\frac{1\;mol\;H2CO3}{2\;mol\;NaHCO3}\;=\;0.0227\;mol\;H2CO3[/tex]
So, the moles of hydrogen carbonate
[tex]0.0227\;mol\;H2CO3.\frac{62.03\;g\;H2CO3}{1\;mol\;H2CO3} = 1.408\;g\;H2CO3[/tex]
Therefore the mass of the product is 1.408 g.
The actual yield of H2CO3 can be determined from mass of sodium bicarbonate and sodium carbonate.
Mass of H2CO3 = Mass of NaHCO3 - Mass of Na2CO3
Mass of H2CO3 = 3.81 g - 2.86 g
Mass of H2CO3 = 0.95 g
Now the percent yield,
% yield
[tex]= \frac{experimental\;yield}{theoretical\;yield}. 100\\= \frac{0.95\;g}{1.408\;g}. 100[/tex]
= 67.47%
≈ 67.5%
Therefore, the percentage yield of the product is 67.5%.
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