Chapter 9, Problem 67A

Interpretation Introduction

Interpretation: The correct statement on the limiting reactant for the below reaction needs to be determined.

  $2C_2{H}_{6(g)}+7O_{2(g)}\stackrel{}{\to }6H_2{O}_{(l)}+4C{O}_{2(g)}$

Concept Introduction: A chemical reaction involves the conversion of reactant to a product by breaking and making of chemical bonds.

There can be more than one reactant molecules involved in a chemical reaction to form the products. The reactant which is present in a limited amount in a chemical reaction is called a limiting reactant.

Answer to Problem 67A

Statement (C) and (D) are always true.

Explanation of Solution

Reason for true statements: The reactant with more mass will have more moles, therefore, it will be present in excess and other reactants will be limiting reactant.

The mole ratio of $C_2{H}_{6(g)}$ and $O_{2(g)}$ is 2:7 thus, the reactant with more mass will have more moles and will be found in excess. It makes another reactant limiting.

Conclusion

Reason for false statements: Statement A and B are false statements.

Explanation: Equal number of mass and moles will not determine the limiting reactant.

The mole ratio of $C_2{H}_{6(g)}$ and $O_{2(g)}$ is 2:7 therefore equal mass or equal moles will not determine the limiting reactant.