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2C4H10 + 13O2 -----> 8CO2 + 10H2O

If 3000 grams of C4H10 react according to this equation, and the actual yield for water is a mass of 1300 grams, what is the percent yield?

User Alyoka
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1 Answer

3 votes

Answer:

27·92 %

Step-by-step explanation:

Given chemical equation is

2C4H10 + 13O2 -----> 8CO2 + 10H2O

So according to the above balanced chemical equation for 2 moles of C4H10, 10 moles of water is produced

Molecular weight of C4H10 is 58 g

Molecular weight of water is 18 g

So for 116 g of C4H10, 180 g of water is produced and therefore for 1 g of C4H10 (180 ÷ 116) g of water is produced

∴ For 1 g of C4H10 1·552 g of water is produced

For 3000 g of C4H10, (3000 × 1·552) g of water is produced

∴ Number of grams of water produced for 3000 g of C4H10 is 4656 g

Percent yield =((experimental amount) ÷ (theoretical amount)) × 100

Here experimental amount is 4656 g

theoretical amount is 1300 g

∴ Percent yield = (1300 ÷ 4656) × 100 = 27·92 %

∴ Percent yield = 27·92 %

User Grant McLean
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