
The mixture of $ N{a_2}C{O_3} $ and $ NaHC{O_3} $ weighs $ 220g $ . Heated precipitate weighs $ 182g $ . What is the percent composition?
Answer
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Hint: When the mixture of $ N{a_2}C{O_3} $ and $ NaHC{O_3} $ were heated, only $ NaHC{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 $ N{a_2}C{O_3} $ and $ NaHC{O_3} $ weighs $ 220g $ . When this mixture is heated, only sodium bicarbonate with the molecular formula of $ NaHC{O_3} $ undergoes decomposition to form water, and carbon dioxide.
The mass of carbon dioxide $ \left( {C{O_2}} \right) $ and water $ \left( {{H_2}O} \right) $ produced is equivalent to the mass of $ {H_2}C{O_3} $
The loss in mass is $ 220g - 182g = 38g $ which is the mass of carbonic acid.
The molar mass of carbonic acid is $ 62gmo{l^{ - 1}} $ .
Thus, the moles of carbonic acid is $ {n_{{H_2}C{O_3}}} = \dfrac{{38g}}{{62gmo{l^{ - 1}}}} = 0.613moles $
From the balanced chemical decomposition,
$ 2NaHC{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 $ {n_{NaHC{O_3}}} = 2 \times 0.163 = 1.225moles $
Mass of sodium bicarbonate is $ {M_{NaHC{O_3}}} = 1.225 \times 84.01 = 102.9g $
The mass of sodium carbonate $ \left( {N{a_2}C{O_3}} \right) $ is the mass of mixture minus mass of sodium bicarbonate.
$ {M_{N{a_2}C{O_3}}} = 220 - 102.9 = 117.1g $
The percent composition of sodium carbonate and sodium bicarbonate will be
$ \% N{a_2}C{O_3} = \dfrac{{117.1g}}{{220}} \times 100 = 53.2\% $
$ \% NaHC{O_3} = \dfrac{{102.9g}}{{220}} \times 100 = 46.8\% $
Thus, the percent of $ NaHC{O_3} $ is $ 46.8\% $ and the percent of $ N{a_2}C{O_3} $ is $ 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.
Complete Step By Step Answer:
Given that the mixture of $ N{a_2}C{O_3} $ and $ NaHC{O_3} $ weighs $ 220g $ . When this mixture is heated, only sodium bicarbonate with the molecular formula of $ NaHC{O_3} $ undergoes decomposition to form water, and carbon dioxide.
The mass of carbon dioxide $ \left( {C{O_2}} \right) $ and water $ \left( {{H_2}O} \right) $ produced is equivalent to the mass of $ {H_2}C{O_3} $
The loss in mass is $ 220g - 182g = 38g $ which is the mass of carbonic acid.
The molar mass of carbonic acid is $ 62gmo{l^{ - 1}} $ .
Thus, the moles of carbonic acid is $ {n_{{H_2}C{O_3}}} = \dfrac{{38g}}{{62gmo{l^{ - 1}}}} = 0.613moles $
From the balanced chemical decomposition,
$ 2NaHC{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 $ {n_{NaHC{O_3}}} = 2 \times 0.163 = 1.225moles $
Mass of sodium bicarbonate is $ {M_{NaHC{O_3}}} = 1.225 \times 84.01 = 102.9g $
The mass of sodium carbonate $ \left( {N{a_2}C{O_3}} \right) $ is the mass of mixture minus mass of sodium bicarbonate.
$ {M_{N{a_2}C{O_3}}} = 220 - 102.9 = 117.1g $
The percent composition of sodium carbonate and sodium bicarbonate will be
$ \% N{a_2}C{O_3} = \dfrac{{117.1g}}{{220}} \times 100 = 53.2\% $
$ \% NaHC{O_3} = \dfrac{{102.9g}}{{220}} \times 100 = 46.8\% $
Thus, the percent of $ NaHC{O_3} $ is $ 46.8\% $ and the percent of $ N{a_2}C{O_3} $ is $ 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.
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