Chapter 9, Problem 19A

Interpretation Introduction

Interpretation:

The mass of CO₂ and CO formed when 5.00 g of carbon is burned under excess and restricted supply of oxygen needs to be deduced from the given reactions.

Concept Introduction:

• A chemical reaction is represented as a chemical equation with the reactants and the products on the left and right side of the reaction arrow respectively.

  $\text{Reactants }\to \text{ Products}$

• The coefficient of a balanced chemical equation, i.e., the stoichiometry gives the number of reactants and products involved in the reaction.
• Chemical equations can therefore be used to determine the amount of products formed from a known quantity of reactants.

Answer to Problem 19A

Mass of CO = 18.33 g

Mass of CO₂ = 11.67 g

Explanation of Solution

a) Reaction in excess oxygen:

The given reaction is:

  ${\text{C(s) + O}}_{\text{2}}\text{(g) }\to {\text{CO}}_2\text{(g) }$

Step 1: Calculate the moles of C present:

Mass of C present = 5.0 g

Atomic weight of C = 12 g/mole

  $\begin{array}{l}\text{Moles of C = }\frac{\text{Mass of C}}{\text{Atomic Mass C}}\\ =\frac{5.0\text{ g}}{12\text{ g/mol}}=0.4167\text{ moles}\end{array}$

Step 2: Calculate the moles of CO₂ formed:

Based on the reaction stoichiometry:

1 mole of Creacts to form 1 mole of CO₂

Therefore, 0.4167 moles of C would produce 0.4167 moles of CO₂

Step 3: Calculate the mass of CO₂ formed:

Moles of CO₂ = 0.4167

Molecular weight of CO2 = 44 g/mol

  $\begin{array}{l}\text{Mass of CO2 = Moles }\times \text{ Molecular weight}\\ =0.4167\text{ moles }\times \text{ 44 g/mol = 18}\text{.33 g}\end{array}$

b) Reaction in restricted oxygen:

The given reaction is:

  $2{\text{C(s) + O}}_{\text{2}}\text{(g) }\to 2\text{CO(g) }$

Step 1: Calculate the moles of C present:

Mass of C present = 5.0 g

Atomic weight of C = 12 g/mole

  $\begin{array}{l}\text{Moles of C = }\frac{\text{Mass of C}}{\text{Atomic Mass C}}\\ =\frac{5.0\text{ g}}{12\text{ g/mol}}=0.4167\text{ moles}\end{array}$

Step 2: Calculate the moles of CO formed:

Based on the reaction stoichiometry:

2 moles of C reacts to form 2 mole of CO

Therefore, 0.4167 moles of C would produce:

  $\frac{0.4167\text{ moles C}\times 2\text{ moles CO}}{2\text{ moles C}}=0.4167\text{ moles CO}$

Step 3: Calculate the mass of CO formed:

Moles of CO = 0.4167

Molecular weight of CO = 28 g/mol

  $\begin{array}{l}\text{Mass of CO = Moles }\times \text{ Molecular weight}\\ =0.4167\text{ moles }\times \text{ 28 g/mol = 11}\text{.67 g}\end{array}$

Conclusion

Therefore, 18.33 g and 11.67 g of CO and CO₂ are produced in excess and restricted supply of oxygen respectively.