   # The equation for the combustion of methanol is 2 CH 3 OH( l ) + 3 O 2 (g) → 2 CO 2 (g) + 4H 2 O( l ). If equal masses of the two reactants are allowed to react, which is the limiting reactant? (a) CH 3 OH (b) O 2 (c) Neither. The reactants are present in the correct stoichiometric ratio. ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 4.2, Problem 1RC
Textbook Problem
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## The equation for the combustion of methanol is 2 CH3OH(l) + 3 O2(g) → 2 CO2(g) + 4H2O(l). If equal masses of the two reactants are allowed to react, which is the limiting reactant? (a) CH3OH (b) O2 (c) Neither. The reactants are present in the correct stoichiometric ratio.

Interpretation Introduction

Interpretation:

The limiting reactant in the given reaction should be determined.

Concept introduction:

• Numberofmole=GivenmassofthesubstanceMolarmass
• The molar mass of an element or compound is the mass in grams of 1 mole of that substance, and it is expressed in the unit of grams per mol (g/mol).
• Limiting reagent: limiting reagent is a reactant, which consumes completely in the chemical reaction. The quantity of the product depends on this limiting reagent, it can be determined with the help of balanced chemical equation for the reaction.
• For chemical reaction balanced chemical reaction equation written in accordance with the Law of conservation of mass.
• Law of conservation of mass states that for a reaction total mass of the reactant and product must be equal.
• Stoichiometric factor is a relationship between reactant and product which is obtained from the balanced chemical equation for a particular reaction.

### Explanation of Solution

Balanced chemical equation for the given reaction is,

2CH3OH(l)+3O2(g)2CO2(g)+4H2O(l)

The amount (moles) of reactants available can be calculated by using the equation

Numberofmole=GivenmassofthesubstanceMolarmass

Considering, the 1.0g mass of reactants each.

1.0gCH3OH×1molCH3OH32.04gCH3OH=0.03121molCH3OH1.0gO2×1molO232.04gO2=0

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