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?

27·92 %

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 %