Calculate Percent Yield Questions: NaNO3 in Impure Substance (URGENT)

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Discussion Overview

The discussion revolves around calculating the percentage yield of NaNO3 in an impure substance after a series of chemical reactions involving Devarda's alloy, H2SO4, and NaOH. The focus is on resolving discrepancies in the calculations presented by the original poster.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • The original poster calculates the percentage yield of NaNO3 as 137%, which raises concerns about the validity of the calculation since the yield cannot exceed 100%.
  • Some participants question the use of the factor 3/2 in the calculation of the amount of NaNO3 reacted, seeking clarification on its derivation.
  • One participant suggests that yields over 100% may indicate issues beyond mere mathematical errors, such as the presence of other nitrates as impurities.
  • Another participant points out an error in the calculation of the amount of NaNO3 reacted, indicating that the original poster's method may be flawed.

Areas of Agreement / Disagreement

Participants express disagreement regarding the calculation steps, particularly the factor used in determining the amount of NaNO3 reacted. There is no consensus on the correct approach or resolution of the percentage yield issue.

Contextual Notes

Participants have not fully resolved the assumptions regarding the nature of the impurities in the original substance, nor have they clarified the derivation of the calculation factors used.

Who May Find This Useful

Students and individuals interested in chemistry calculations, particularly those dealing with percentage yields and reaction stoichiometry, may find this discussion relevant.

Originaltitle
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% yield questions (URGENT)

Homework Statement


We have 3 equations:
2w1tzcy.jpg

1.64 g of an impure NaNO3-containing substance is reacted with Devarda's alloy. The amount of NH3 got from this reaction is reacted with 25cm3 1.00 moldm-3 H2SO4. The H2SO4 left over is reacted with 16.2 cm3 2.00 moldm-3 NaOH. Calculate the percentage yield of NaNO3 in the impure substance.2. The attempt at a solution

My attempt at an answer:
1. Amount of H2SO4 reacted with NaOH = (2.00 x 16.2 x 10-3) / 2 = 0.0162 moles.
2. Amount of H2SO4 reacted with NH3 = 0.025 - 0.0162 = 0.0088 moles.
3. Amount of NH3 reacted = (0.0088 x 2) = 0.0176 moles.
4. Amount of NaNO3 reacted = 0.0176 x (3/2) = 0.0264.
5. Mass of NaNO3 reacted = 0.0264 x 85 = 2.244 g.

% yield = 2.244/1.64 = 137 %.

It's wrong because the final mass can't be more than the initial. HELP!
 
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Originaltitle said:
4. Amount of NaNO3 reacted = 0.0176 x (3/2) = 0.0264.

Why 3/2?

It's wrong because the final mass can't be more than the initial. HELP!

% yields over 100% do happen. They usually mean something is wrong, but it is not necessarily a math error. For example in this case it could mean that the impurity is some other nitrate.
 
The error is in the line:
Originaltitle said:
4. Amount of NaNO3 reacted = 0.0176 x (3/2) = 0.0264.
 
Edit: But Borek already told you that...
 

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