Chapter 9, Problem 99CP

Introductory Chemistry: A Foundati...

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

Chapter
Section

Introductory Chemistry: A Foundati...

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

Sulfur dioxide gas reacts with sodium hydroxide to form sodium sulfite and water. The unbalanced chemical equation for this reaction is as follows::math> SO 2 ( g ) + NaOH ( s ) → Na 2 SO 3 ( s ) + H 2 O ( l ) suming you react 38.3 g of sulfur dioxide with 32.8 g of sodium hydroxide and assuming that the reaction goes to completion, calculate the mass of each product formed.

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

The mass of sulphur dioxide is 38.3 g and that of sodium hydroxide is 32.8 g.

The balanced chemical reaction is as follows:

SO2(g)+2NaOH(s)Na2SO3(s)+H2O(l)

To calculate the mass of each product, first determine the limiting reactant. Calculate the number of moles of sulphur dioxide and sodium hydroxide as follows:

n=mM

Molar mass of sulphur dioxide is 64.066 g/mol, number of moles will be:

n=38.3 g64.066 g/mol=0.598 mol

Since, 1 mol of sulphur dioxide gives 1 mol of Na2SO3 thus, 0.598 mol of sulphur dioxide gives 0.598 mol of Na2SO3.

Molar mass of Na2SO3 is 126.043 g/mol thus, mass of Na2SO3 will be:

m=n×M

Putting the values,

m=(0.598 mol)(126.043 g/mol)=75.35 g

Now, calculate number of moles of NaOH as follows:

n=mM

Molar mass of NaOH is 39.997 g/mol thus, number of moles will be:

n=32.8 g39.997 g/mol=0

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