   # Hydrogen cyanide is produced industrially from the reaction of gaseous ammonia, oxygen, and methane: 2 NH 3 ( g ) + 3 O 2 ( g ) + 2 CH 4 ( g ) → 2 HCN ( g ) + 6 H 2 O ( g ) If 5.00 × 10 3 kg each of NH 3 , O 2 , and CH 4 are reacted, what mass of HCN and of H 2 O will be produced, assuming 100% yield? ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 5, Problem 121E
Textbook Problem
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## Hydrogen cyanide is produced industrially from the reaction of gaseous ammonia, oxygen, and methane: 2 NH 3 ( g )   +   3 O 2 ( g )   +   2 CH 4 ( g ) → 2 HCN ( g )   +   6 H 2 O ( g ) If 5.00 × 103 kg each of NH3, O2, and CH4 are reacted, what mass of HCN and of H2O will be produced, assuming 100% yield?

Interpretation Introduction

Interpretation: The amount of HCN and H2O that can be produced from the stated reaction is to be calculated.

Concept introduction: The mass of a substance can be obtained by using the number of moles of the substance present and its molar mass. The formula used to calculate the mass of a given substance is,

Massofthesubstance=(Numberofmoles)×(Molarmassofthesubstance)

To determine: The mass of HCN and H2O that can be produced from the stated reaction.

### Explanation of Solution

To determine: The mass of HCN that can be produced from the stated reaction.

Given

The stated chemical reaction is,

2NH3(g)+3O2(g)+2CH4(g)2HCN(g)+6H2O(g)

The given mass of NH3 is 5.00×103kg(5.00×106g) .

The given mass of O2 is 5.00×103kg(5.00×106g) .

The given mass of CH4 is 5.00×103kg(5.00×106g) .

The yield of the reaction is 100% .

The molar mass of NH3 =N+3H=(14+(3×1))g/mol=17g/mol

The molar mass of O2 =2O=(2×16)g/mol=32g/mol

The molar mass of CH4 =C+4H=(12+(4×1))g/mol=16g/mol

Formula

The number of moles of a substance is calculated by the formula,

Numberofmoles=GivenmassofthesubstanceMolarmassofthesubstance

Substitute the value of the given mass and the molar mass of NH3 , O2 and CH4 in the above expression.

NumberofmolesofNH3=5×106g17g/mol=0.294×106mol

NumberofmolesofO2=5×106g32g/mol=0.156×106mol

NumberofmolesofCH4=5×106g16g/mol=0.312×106mol

According to the stated reaction,

2mol of NH3 react with 2mol of CH4 .

1mol of NH3 reacts with 1mol of CH4 .

0.294×106mol of NH3 react with 0.294×106mol of CH4 .

The number of moles of CH4 given are 0.312×106mol . Hence, it is not the limiting reactant.

According to the stated reaction,

2mol of NH3 react with 3mol of O2 .

1mol of NH3 reacts with O2 =(32)mol

0.294×106mol of NH3 reacts with O2 =(3×0

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