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Question: The mixture of \( N{a_2}C{O_3} \) and \( NaHC{O_3} \) weighs \( 220g \) . Heated precipitate weighs ...

The mixture of Na2CO3N{a_2}C{O_3} and NaHCO3NaHC{O_3} weighs 220g220g . Heated precipitate weighs 182g182g . What is the percent composition?

Explanation

Solution

When the mixture of Na2CO3N{a_2}C{O_3} and NaHCO3NaHC{O_3} were heated, only NaHCO3NaHC{O_3} decomposes to form sodium carbonate, carbon dioxide, and water. The mass of water and carbon dioxide is equivalent to carbonic acid whose moles can be calculated from heated precipitate and molar mass of carbonic acid, these moles used to determine the percent composition of mixture.

Complete Step By Step Answer:
Given that the mixture of Na2CO3N{a_2}C{O_3} and NaHCO3NaHC{O_3} weighs 220g220g . When this mixture is heated, only sodium bicarbonate with the molecular formula of NaHCO3NaHC{O_3} undergoes decomposition to form water, and carbon dioxide.
The mass of carbon dioxide (CO2)\left( {C{O_2}} \right) and water (H2O)\left( {{H_2}O} \right) produced is equivalent to the mass of H2CO3{H_2}C{O_3}
The loss in mass is 220g182g=38g220g - 182g = 38g which is the mass of carbonic acid.
The molar mass of carbonic acid is 62gmol162gmo{l^{ - 1}} .
Thus, the moles of carbonic acid is nH2CO3=38g62gmol1=0.613moles{n_{{H_2}C{O_3}}} = \dfrac{{38g}}{{62gmo{l^{ - 1}}}} = 0.613moles
From the balanced chemical decomposition,
2NaHCO3Na2CO3+CO2+H2O2NaHC{O_3} \to N{a_2}C{O_3} + C{O_2} + {H_2}O
Two moles of sodium bicarbonate produce one mole of carbonic acid.
Thus, the moles of sodium bicarbonate is nNaHCO3=2×0.163=1.225moles{n_{NaHC{O_3}}} = 2 \times 0.163 = 1.225moles
Mass of sodium bicarbonate is MNaHCO3=1.225×84.01=102.9g{M_{NaHC{O_3}}} = 1.225 \times 84.01 = 102.9g
The mass of sodium carbonate (Na2CO3)\left( {N{a_2}C{O_3}} \right) is the mass of mixture minus mass of sodium bicarbonate.
MNa2CO3=220102.9=117.1g{M_{N{a_2}C{O_3}}} = 220 - 102.9 = 117.1g
The percent composition of sodium carbonate and sodium bicarbonate will be
%Na2CO3=117.1g220×100=53.2%\% N{a_2}C{O_3} = \dfrac{{117.1g}}{{220}} \times 100 = 53.2\%
%NaHCO3=102.9g220×100=46.8%\% NaHC{O_3} = \dfrac{{102.9g}}{{220}} \times 100 = 46.8\%
Thus, the percent of NaHCO3NaHC{O_3} is 46.8%46.8\% and the percent of Na2CO3N{a_2}C{O_3} is 53.2%53.2\% .

Note:
Sodium bicarbonate is made up of sodium, hydrogen, and acid. But sodium carbonate is made up of sodium, and acid. It is formed by the combination of sodium hydroxide, a strong base and weak acid. Thus, it remains as a solid when the mixture of sodium bicarbonate and sodium carbonate is heated.