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Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425

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BuyFindarrow_forward

Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425
Textbook Problem
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Ammonia gas reacts with sodium metal to form sodium amide ( NaNH 2 ) and hydrogen gas. The unbalanced chemical equation for this reaction is as follows:

msp;  NH 3 ( g ) + Na 2 ( s ) NaNH 2 ( g ) + H 2 ( g )

suming that you start with 32.8 g of ammonia gas and 16.6 g of sodium metal and assuming that the reaction goes to completion, determine the mass (in grams) of each product.

Interpretation Introduction

Interpretation:

The mass of each product formed should be calculated.

Concept Introduction:

The expected or theoretical yield of the product depends on the limiting reactant of the reaction. Limiting reactant is the reactant that limits the mass of product formed in the reaction. The reactant other than limiting reactant is known as excess reactant.

Explanation

To calculate the amount of product, limiting reactant should be determined. The mass of ammonia gas is 32.8 g and that of sodium metal is 16.6 g. Calculate number of moles of ammonia gas and sodium metal as follows:

Molar mass of ammonia gas is 17.031g/mol, number of moles will be:

n=mM

Putting the values,

n=32.8 g17.031 g/mol=1.925 mol

The balanced chemical reaction is as follows:

2NH3(g)+2Na(s)2NaNH2(s)+H2(g)

2 moles of ammonia gives 2 mol of NaNH2 or 1 mol of ammonia gives 1 mol of NaNH2 thus, 1.925 mol of ammonia gives 1.925 mol of NaNH2

Molar mass of NaNH2 is 39.01 g/mol thus, mass of NaNH2 will be:

m=n×M

Putting the values,

m=(1.925 mol)(39.01 g/mol)=75.13 g

Similarly, calculate mass of NaNH2 from sodium metal as follows:

First calculate the number of moles of sodium metal as follows:

n=mM

Molar mass of sodium is 23 g/mol putting the values,

n=16

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