A 2.556g sample of baking soda (NaHCO3)decomposes to produce 0.876g of solid sodium carbonate. Caculate the theoretical yield and the percent yield of the sodium carbonate. The equation is:
2NaHCO3(s) -> Na2CO3(s) + H2O + CO2(g)
This is a straight forward stoichiometry problem.
2NaHCO3 ==> Na2CO3 + CO2 + H2O
mols NaHCO3 = grams/molar mass = approx (estimated) 0.03
Convert to mols Na2CO3. The ratio in the equation is 2:1; therefore, mols Na2CO3 formed is estimated 0.015
grams Na2CO3 formed is g = mols x molar mass = estimated 1.6 g. This is the theoretical yield (TY)
Actual yield (AY) in the problem is 0.876g
% yield = (AY/TY)*100 = >
Check my work.
thank you, i forgot to convert the mols, once i put that part in. I got about the same thing as you
To calculate the theoretical yield of sodium carbonate (Na2CO3), we need to determine the molar ratio between baking soda (NaHCO3) and sodium carbonate (Na2CO3), and then use the given mass of baking soda (NaHCO3) to calculate the theoretical yield.
1. Calculate the molar mass of NaHCO3:
Na: 1 atom x 22.99 g/mol = 22.99 g/mol
H: 1 atom x 1.01 g/mol = 1.01 g/mol
C: 1 atom x 12.01 g/mol = 12.01 g/mol
O: 3 atoms x 16.00 g/mol = 48.00 g/mol
Total molar mass = 22.99 + 1.01 + 12.01 + 48.00 = 83.01 g/mol
2. Calculate the number of moles of NaHCO3:
Moles = Mass / Molar mass
Moles = 2.556 g / 83.01 g/mol ≈ 0.03081 mol
3. Based on the balanced equation, the molar ratio between NaHCO3 and Na2CO3 is 2:1.
Therefore, moles of Na2CO3 = 0.03081 mol / 2 = 0.01541 mol
4. Calculate the mass of Na2CO3 using the molar mass:
Molar mass of Na2CO3:
Na: 2 atoms x 22.99 g/mol = 45.98 g/mol
C: 1 atom x 12.01 g/mol = 12.01 g/mol
O: 3 atoms x 16.00 g/mol = 48.00 g/mol
Total molar mass = 45.98 + 12.01 + 48.00 = 105.99 g/mol
Mass of Na2CO3 = Moles x Molar mass
Mass of Na2CO3 = 0.01541 mol x 105.99 g/mol ≈ 1.635 g
Therefore, the theoretical yield of sodium carbonate (Na2CO3) is approximately 1.635 g.
To calculate the percent yield, we need to compare the actual yield (0.876 g) with the theoretical yield (1.635 g) and calculate the percentage.
Percent yield = (Actual yield / Theoretical yield) x 100%
Percent yield = (0.876 g / 1.635 g) x 100% ≈ 53.6%
Therefore, the percent yield of the sodium carbonate is approximately 53.6%.
To calculate the theoretical yield of sodium carbonate (Na2CO3), you first need to determine the balanced chemical equation and the molar ratios between the reactants and products.
According to the given equation:
2NaHCO3(s) -> Na2CO3(s) + H2O + CO2(g)
From the equation, you can see that 2 moles of sodium bicarbonate (NaHCO3) will produce 1 mole of sodium carbonate (Na2CO3). Therefore, the molar ratio between Na2CO3 and NaHCO3 is 1:2.
Now, let's calculate the moles of NaHCO3 using the given mass:
molar mass of NaHCO3 = 22.99 g/mol (Na) + 1.01 g/mol (H) + 12.01 g/mol (C) + 3(16.00 g/mol) (O) = 84.01 g/mol
moles of NaHCO3 = mass of NaHCO3 / molar mass of NaHCO3
= 2.556 g / 84.01 g/mol
≈ 0.030 g/mol (rounded to three significant figures)
Since the molar ratio of Na2CO3 to NaHCO3 is 1:2, the moles of Na2CO3 produced will be half of the moles of NaHCO3:
moles of Na2CO3 = moles of NaHCO3 / 2
≈ 0.015 mol (rounded to three significant figures)
Now, we can calculate the theoretical yield of Na2CO3 using the molar mass of Na2CO3:
molar mass of Na2CO3 = 22.99 g/mol (Na) + 12.01 g/mol (C) + 3(16.00 g/mol) (O)
= 105.99 g/mol
theoretical yield of Na2CO3 = moles of Na2CO3 * molar mass of Na2CO3
= 0.015 mol * 105.99 g/mol
≈ 1.59 g (rounded to three significant figures)
Therefore, the theoretical yield of sodium carbonate is approximately 1.59 grams.
To calculate the percent yield, you need to compare the actual yield (0.876g) with the theoretical yield (1.59g) and calculate the percentage:
percent yield = (actual yield / theoretical yield) * 100
= (0.876 g / 1.59 g) * 100
≈ 55.1% (rounded to one decimal place)
Therefore, the percent yield of sodium carbonate is approximately 55.1%.