   # The reaction of methane and water is one way to prepare hydrogen for use as a fuel: CH 4 (g) + H 2 O(g) → CO(g) + 3 H 2 (g) If this reaction has a 37% yield under certain conditions, what mass of CH 4 is required to produce 15 g of H 2 ? ### 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, Problem 23PS
Textbook Problem
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## The reaction of methane and water is one way to prepare hydrogen for use as a fuel:CH4(g) + H2O(g) → CO(g) + 3 H2(g)If this reaction has a 37% yield under certain conditions, what mass of CH4 is required to produce 15 g of H2?

Interpretation Introduction

Interpretation:

It should be determine that the mass of CH4 required to produce 15g of H2 if the reaction has 37% yield.

Concept introduction:

• The relation between the number of moles and mass of the substance is ,

Numberofmole=MassingramMolarmass

Mass in gramofthe substance = Numberofmole×Molarmass

• 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).
• 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.
• Amountof substance=Concentrationofsubstance×volumeofthesubstance
• Mass percent of elements of a compound is the ratio of mass of element to the mass of whole compound and multiplied with hundred.

### Explanation of Solution

Balanced chemical equation for the given reaction is,

CH4(g)+H2O(g)CO(g)+3H2(g)

From the balanced chemical equation it is clear that H2 and CH4 react in a 3:1ratio.  From the mass percent of reaction actual mass of H2 involved in this reaction can be calculated.

Thus,

37%of15gH2=150.37=40.54gH2

The amount (moles) of H2 available can be determined as follows,

Amount of H2= 40.54gH22

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