How do I find the theoretical yield in moles, in grams, and the percent yield?

[匿名]

2016-10-14 6:41 am

If I have an equation that looks like this:

2NaHCO3 (s) --> Na2CO3 (s) + CO2 (g) + H2O (l)

and I got, as the mass of NaHCO3, 4.221 g, and as the mass of Na2CO3 produced, 2.504 g, then how could I find the theoretical yield of Na2CO3 in both moles and grams? How could I find the percent yield of Na2CO3?

I would really appreciate if someone could walk me through the steps, as I really just want to understand how to do it.

更新1:

Also, if it's not 100%, why would the yield be less than or more than that? Is it because of human error in the lab?

更新2:

Wait, so why am I not using the mass of the Na2CO3 produced? Does that not show up in the equations?

更新3:

Never mind. Thank you!!!!!!!!

回答 (1)

wanszeto

2016-10-14 6:58 am

✔ 最佳答案

Molar mass of NaHCO₃ = (22.99 + 1.01 + 12.01 + 16.00×3) g/mol = 84.01 g/mol
Molar mass of Na₂CO₃ = (22.99×2 + 12.01 + 16.00×3) g/mol = 105.99 g/mol

2NaHCO₃(s) → Na₂CO₃(s) + CO₂(g) + H₂O(l)
OR: Mole ratio NaHCO₃ : Na₂CO₃ = 2 : 1

No. of moles of NaHCO₃ reacted = (4.221 g) / (84.01 g/mol) = 0.05024 mol
Theoretically, no. of moles of Na₂CO₃ formed = (0.05024 mol) × (1/2) = 0.02512 mol
Theoretical yield of Na₂CO₃ = (105.99 g/mol) × (0.02512 mol) = 2.662 g

Percent yield of Na₂CO₃ = (2.504/2.662) × 100% = 94.06%

The percent yield is less than 100% because some product is lost during experiment steps.

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