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Aloiza [94]
3 years ago
9

A reaction between solid sulphur and oxygen produces sulfur dioxide. What is the percent yield for a reaction if you predicted t

he formation of 680g of sulfur dioxide and actually recovered only 384g?
Chemistry
1 answer:
Oduvanchick [21]3 years ago
3 0

Answer:

56.47%

Explanation:

In this question, we are given;

  • Theoretical yield of sulfur dioxide as 680 g
  • Actual yield of sulfur dioxide as 384 g

We are required to calculate the percent yield for the reaction;

  • We need to know how percent yield is calculated;
  • % yield = (Actual yield ÷ Theoretical yield) × 100%

Therefore;

In this case;

% yield = (384 g ÷ 680 g) × 100%

            = 56.47%

Thus, the percent yield for the reaction is 56.47%

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For 1 mol of N2, we need 3 moles of H2 to produce 2 moles of NH3

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The 30 moles will be completely be consumed.

N2 is in excess. There will react 30/3 =10 moles

There will remain 30 -10 = 20 moles (this in the case of a 100% yield)

In a 50 % yield, there will remain 20 + 0,5*10 = 25 moles. there will react 5 moles.

<u>Step 4:</u> Calculate moles of NH3

There will be produced, 30/ (3/2) = 20 moles of NH3 (In case of 100% yield)

For a 50% yield there will be produced, 10 moles of NH3

<u>Step 5</u>: Calculate the mass of NH3

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Mass of NH3 = 340.6 grams = theoretical yield ( 100% yield)

<u>Step 6: </u>Calculate actual mass

50% yield = actual mass / theoretical mass

actual mass = 0.5 * 340.6

actual mass = 170.3 grams

<u>Step 7:</u> The mass of nitrogen remaining

There remain 20 moles of nitrogen + 50% of 10 moles = 25 moles remain

Mass of nitrogen = 25 moles * 28 g/mol

Mass of nitrogen = 700 grams

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