Chemistry Homework

Question

Carbon monoxide reacts with oxygen gas to form CO₂, as shown by the balanced equation below.

A reaction vessel contains 2.700 g of CO and 2.700 g of O₂.

How many moles of CO and O₂ do these masses represent?

What is the maximum number of moles of CO₂ recovered, based on the number of moles of the given reactants?

Answer 1

On this question, first, we need a balanced equation:

2 CO_(g) + O_{2 (g)} → 2 CO_{2 (g)}

After that, since you are given amounts of both reactants, CO and O₂, we need to determine which will limit the amount of product formed (which is the limiting reactant). This is done by converting the masses of each of the reactants into moles of reactant (using molar mass), then converting to moles of CO₂ (using the stoichiometric coefficients from the balanced equation).

2.700 g CO (1 mol CO / 28.010 g CO) = 0.09639 mol CO (2 mol CO₂ / 2 mol CO) = 0.09639 mol CO₂

2.700 g O₂ (1 mol O₂ / 31.998 g O₂) = 0.08438 mol O₂ (2 mol CO₂ / 1 mol O₂) = 0.1688 mol CO₂

The limiting reactant is the reactant that will be used up first. As such, it will limit the amount of product that can maximally be produced. We can determine which is the limiting reactant by seeing which results in the smaller quantity of product (usually in moles as it saves a step). From the work above, we see that CO produces a smaller quantity of moles of CO₂, and is, therefore, the limiting reactant. As such, the theoretical yield is 0.09639 moles of CO₂. The theoretical yield is the maximum amount of CO₂ that can be recovered.

As a note, the theoretical yield can also be given in grams. If the question asks for the maximum amount of grams of CO₂ that can be made, you would take the theoretical yield in moles and multiply It by the molar mass:

0.09639 mol CO₂ (44.009 g CO₂ / 1 mol CO₂) = 4.242 g CO₂