Question

How many grams of butane (C H20) must be burned in an excess of O, to produce 15.0
g of CO2? NOTE: Butane reacts with oxygen gas to produce carbon dioxide and water.

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

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

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

$= 1.87 \: g$

Answer 2

The grams of butane must be burned in an excess of Oxygen, to produce 15g of CO₂ is 1.87 grams.

What is the relation between mass & moles?

Relation between mass and moles will be represented as:

n = W/M, where

W = require or given mass

M = molar mass

Given balance chemical reaction is:

C₄H₁₀ + 13/2O₂ → 4CO₂ + 5H₂O

Moles of 15g of CO₂ = 15g / 44g/mol = 0.34 moles

From the stoichiometry of the reaction, it is clear that:

4 moles of CO₂ = produced by 1 mole of C₄H₁₀

0.34 moles of CO₂ = produced by 1/4×0.34=0.085 moles of C₄H₁₀

Now we convert moles of butane to mass as:

W = (0.085)(22) = 1.87 g

Hence required mass of butane is 1.87 grams.

To know more about mass & moles, visit the below link:
https://brainly.com/question/24639749