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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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he production capacity for acrylonitrile ( C 3 H 3 N ) in the United States is over 2 billion pounds per year. Acrylonitrile, the building block tor acrylonitrile fibers and a variety of plastics, is produced from gaseous propylene, ammonia, and oxygen:

:math> 2 C 3 H 6 ( g ) + 2 NH 3 ( g ) + 3 O 2 2 C 3 H 3 N ( g ) + 6 H 2 O ( g )

l type='a'>

  • Assuming 100% yield, determine the mass of acrylonitrile which can be produced from the mixture below:
  • l>

    Mass Reactant

    msp;  5.23 × 10 2 g propylene

    td>

    msp;  5 .00 × 10 2 g ammonia

    td>

    msp;  1 .00 × 10 3 g oxygen

    td>

    i>What mass of water is formed from your mixture?

  • Calculate the mass (in grams) of each reactant after the reaction is complete.
  • Interpretation Introduction

    (a)

    Interpretation:

    The mass of acrylonitrile which can be produced from the reactants given should be determined.

    Concept Introduction:

    To determine how much product can be formed from a given mixture of reactants, we have to look for the reactant that is limiting; the one that runs out first and thus limits the amount of product that can form. The reactant that runs out first limiting the amount of products form is called the limiting reactant or limiting reagent.

    Explanation

    Number of moles of C3 H6 = 5.23×102 g42.078 g/mol = 12.4 mol

    Number of moles of NH3 = 5×102 g17.034 g/mol = 29.4 mol

    Number of moles of O2 = 1×103 g32.00 g/mol = 31.3 mol

    Possibility I: if C3 H6 runs out first

    Balanced equation

    2C3H6(g)    +        2NH3(g)    +      3O2(g)            2C3H3N(g)      +       6H2O(g)

    Before

    12.4 mol

    29.4 mol

    31.3 mol

    0 0

    Change

    12.4 mol

    12.4 mol

    18.6 mol

    +12.4 mol

    +37.2 mol

    ______________________________________________________________________________

    After

    0 17.0 mol

    12.7 mol 12.4 mol

    37.2 mol

    Possibility II: if NH3 runs out first

    Balanced equation

    2C3H6(g)    +        2NH3(g)    +      3O2(g)            2C3H3N(g)      +       6H2O(g)

    Before

    12

    Interpretation Introduction

    (b)

    Interpretation:

    Mass of water formed from the mixture should be calculated.

    Concept Introduction:

    Molar mass=Mass of the sample in gramsNumber of moles

    Mass of the sample in grams = Number of moles × Molar mass.

    Interpretation Introduction

    (c)

    Interpretation:

    Mass of each reactant after the reaction is complete should be calculated.

    Concept Introduction:

    Molar mass=Mass of the sample in gramsNumber of moles

    Mass of the sample in grams = Number of moles × Molar mass.

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