a. Balanced equation for the decomposition of sodium bicarbonate by heating is:
2 NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g)
b. To heat the mixture, the student can use a Bunsen burner or a hot plate in a fume hood to prevent the release of any harmful gases. The mixture should be heated until there is no more mass loss. To determine the mass loss, the student can weigh the vial and its contents before and after heating. The difference between the two weights will be the mass loss.
c. The student can know that all the sodium bicarbonate has been decomposed when there is no more mass loss. The heating should be continued until the mass of the vial and its contents remains constant.
d. The mass loss of 2.28 g is due to the decomposition of 1 mol of NaHCO3, which has a molar mass of 84 g/mol. Therefore, the original mixture contained 2.28/84 = 0.027 mol of NaHCO3. The mass of Na2CO3 in the mixture can be calculated as follows:
mass of Na2CO3 = mass of mixture - mass loss = 14.00 g - 2.28 g = 11.72 g
The molar mass of Na2CO3 is 106 g/mol, so the number of moles of Na2CO3 in the mixture is:
moles of Na2CO3 = mass of Na2CO3/molar mass of Na2CO3 = 11.72 g/106 g/mol = 0.111 mol
The percentage of Na2CO3 in the original mixture is:
% Na2CO3 = (moles of Na2CO3/total moles of carbonate) x 100% = (0.111/0.138) x 100% = 80.4%
e. To determine which is the limiting reactant, we need to compare the number of moles of NaHCO3 and Na2CO3 in the mixture. We know that the mixture contained 0.027 mol of NaHCO3 and 0.111 mol of Na2CO3. Since the balanced equation shows that 2 moles of NaHCO3 produce 1 mole of Na2CO3, the number of moles of Na2CO3 that can be produced from 0.027 mol of NaHCO3 is:
moles of Na2CO3 = 0.027 mol NaHCO3 x (1 mol Na2CO3/2 mol NaHCO3) = 0.0135 mol Na2CO3
Since the actual amount of Na2CO3 in the mixture is greater than 0.0135 mol, we can conclude that NaHCO3 is the limiting reactant, and Na2CO3 is in excess.
d.
Mass of NaHCO3 = 2.28 g / 84 g/mol = 0.027 mol NaHCO3
Mass of Na2CO3 = 14.00 g - 2.28 g = 11.72 g
Moles of Na2CO3 = 11.72 g / 106 g/mol = 0.111 mol Na2CO3
Total moles of carbonate = moles of NaHCO3 + moles of Na2CO3 = 0.027 mol + 0.111 mol = 0.138 mol
% Na2CO3 = (moles of Na2CO3 / total moles of carbonate) x 100% = (0.111 mol / 0.138 mol) x 100% = 80.4%
e.
Moles of Na2CO3 that can be produced from 0.027 mol of NaHCO3 = 0.027 mol NaHCO3 x (1 mol Na2CO3 / 2 mol NaHCO3) = 0.0135 mol Na2CO3
Since the actual amount of Na2CO3 in the mixture (0.111 mol) is greater than 0.0135 mol, NaHCO3 is the limiting reactant, and Na2CO3 is in excess.
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