# Chemistry help?

Given the following equation:

2C4H10+13O2—>8CO2+10H20

How many grams of the C4H10 are needed to produce 26.00 g of H2O

Relevance
• (26.00 g H2O) / (18.01532 g H2O/mol) x (2 mol C4H10 / 10 mol H2O) x

(58.1222 g C4H10/mol) = 16.78 g C4H10

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• Butane + Oxygen ➜ Carbon Dioxide + water

2C₄H₁₀ + 13O₂ ➜ 8CO₂ + 10H₂O

molecular weights

C = 12

H = 1

O = 16

2C₄H₁₀ = 2•58 = 116

13O₂ = 13•32 = 416

8CO₂ = 8•44 = 352

10H₂O = 10•18 = 180

check 116+416 = 352+180 = 532

116 grams of C₄H₁₀ + 416 grams of O₂ ➜ 352 grams of CO₂ + 180 grams of H₂O

2 mole of C₄H₁₀ + 13 moles of O₂ ➜ 8 mole of CO₂ + 10 moles of H₂O

How many grams of the C4H10 are needed to produce 26.00 g of H2O

ratio of C₄H₁₀ to H₂O is 116/180

116/180 = x/26

x = 16.8 g

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• Anonymous
3 months ago

26 g of water is about 1.5 moles.  Moles of butane = 1.5 moles H2O * 2 moles butane / 10 moles water.  Multiply moles of butane by its molar mass to get mass.

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