Question

How many grams of butane (C H20) must be burned in an excess of O, to produce 15.0

Answer 1

Answer:

Mass of butane = 1.87 g

Explanation:

Form a balanced chemical equation for the reaction stated:

C₄H₁₀ + 13/2 O₂ → 4CO₂ + 5H₂O

Number of moles of CO₂

$= \frac{mass}{molar \: mass}$

$= \frac{15}{12 + 2(16)}$

$= 0.34 \: mol$

From the equation,

1 mol C₄H₁₀ : 4 mol CO₂

0.085 mol C₄H₁₀ : 0.34 mol CO₂

Mass of butane

$= number \: of \: moles \times molar \: mass$

$= 0.085 \times (12 + 10(1))$

$= 1.87 \: g$