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Step2247 [10]
3 years ago
6

Consider the reaction below. If you start with 4.00 moles of C3H8 (propane) and 4.00 moles of O2 , how many moles of propane wou

ld be consumed?
Chemistry
1 answer:
fgiga [73]3 years ago
7 0

Answer:

0.800 mol

Explanation:

We have the amounts of two reactants, so this is a limiting reactant problem.  

We know that we will need a balanced equation with moles of the compounds involved.  

Step 1. <em>Gather all the information</em> in one place.

            C₃H₈ + 5O₂ ⟶ 3CO₂ + 4H₂O

n/mol:  4.00    4.00

===============

Step 2. Identify the <em>limiting reactant </em>

Calculate the <em>moles of CO₂</em> we can obtain from each reactant.  

<em>From C₃H₈:</em>

The molar ratio of CO₂: C₃H₈ is 3:1

Moles of CO₂ = 4.00 × 3/1

Moles of CO₂ = 12.0 mol CO₂

<em>From O₂</em>:

The molar ratio of CO₂: O₂ is 3:5.

Moles of CO₂ = 4.00 × ⅗

Moles of CO₂ = 2.40 mol CO₂

O₂ is the limiting reactant because it gives the smaller amount of CO₂.

==============

Step 3. Calculate the <em>moles of C₃H₈ consumed</em>.

The molar ratio of C₃H₈:O₂ is 1:5.

Moles of C₃H₈ = 4.00 × ⅕

Moles of C₃H₈ = 0.800 mol C₃H₈

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