C4H10+O2=CO2+H2O

Question

If the reaction produces 90 grams of water, how much butane was used?

ive found the answer using calculators, but I cannot find an explanation. I really need to know how the answer was derived. Please help.

Answer 1

the balanced chemical equation for this reaction is

2 C₄​H₁₀ + 13 O₂ ​→ 8 CO₂​ + 10 H₂​O

meaning that 2 moles of butane (C₄​H₁₀) produces 10 moles of water (H₂​O)

so first you want to take the 90 grams of water and convert that into moles

90 g / (18.016 g/mol) = 4.996 mol H₂O

then you can convert that into moles of butane by multiplying by the molar ratio

4.996 mol H₂O x (2 moles C₄​H₁₀ / 10 moles H₂​O) = 0.999 mol butane

finally then i would scale that by the molar mass to convert the balance into grams

0.999 mol butane x (58.12 g/mol) = 58.07 g

meaning that 58.07 grams of butane were used in this scenario

Answer 2

Find an answer here.

https://youtu.be/45mKzpJ5iIg