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

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SUMMARY

The discussion centers on calculating the percentage yield of NaNO3 from an impure substance weighing 1.64 g, reacted with Devarda's alloy and sulfuric acid (H2SO4). The calculations initially led to a yield of 137%, indicating a potential error in the assumptions or calculations. The critical mistake identified was in the calculation of the amount of NaNO3 reacted, specifically the use of a 3/2 ratio without proper justification. The presence of impurities, such as other nitrates, may also contribute to yields exceeding 100%.

PREREQUISITES
  • Understanding of stoichiometry and mole calculations
  • Familiarity with acid-base reactions, specifically involving H2SO4 and NaOH
  • Knowledge of Devarda's alloy and its role in ammonia production
  • Basic principles of percentage yield in chemical reactions
NEXT STEPS
  • Review stoichiometric calculations in chemical reactions
  • Learn about the role of impurities in yield calculations
  • Study the properties and reactions of Devarda's alloy
  • Explore common mistakes in percentage yield calculations
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Chemistry students, educators, and professionals involved in analytical chemistry or laboratory work, particularly those focused on yield calculations and reaction stoichiometry.

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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