Question

A reaction vessel contains 10.9 g of CO and 10.9 g of O2. How many grams of CO2 could be produced according to the following unbalanced reaction?

a) 22.7 g
b) 21.8 g
c) 20.5 g
d) 19.6 g

Answer 1

Using stoichiometry, the calculation shows that the reaction of 10.9 g of O2 with excess CO would theoretically produce 29.9 g of CO2. However, this value does not match any of the provided options, indicating a possible error in the question or the need for additional information.

Step-by-step explanation:

The subject of this question is stoichiometry, a branch of Chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. Given are 10.9 g of CO and 10.9 g of O2, and we need to find out how many grams of CO2 could be produced from these reactants according to a balanced chemical equation.

First, we need to balance the chemical equation:

• 2CO + O2 → 2CO2

Now we can calculate the moles of CO and O2:

• Molar mass of CO = 28.01 g/mol and molar mass of O2 = 32.00 g/mol
• Moles of CO = 10.9 g / 28.01 g/mol = 0.389 moles
• Moles of O2 = 10.9 g / 32.00 g/mol = 0.340 moles

According to the balanced equation, it takes 1 mole of O2 to react with 2 moles of CO. Because we have more moles of CO than required for the moles of O2 we have, O2 is the limiting reactant. Thus, the reaction is limited by the amount of O2, which means that we will use all 0.340 moles of O2 to produce CO2.

From the balanced equation, 1 mole of O2 produce 2 moles of CO2, so:

• Moles of CO2 = 2 × 0.340 moles = 0.680 moles

Molar mass of CO2 = 44.01 g/mol, so:

• Grams of CO2 = 0.680 moles × 44.01 g/mol = 29.9 g of CO2 (theoretical yield)

The options given do not include the correct answer which is 29.9 g; therefore, the question may contain a mistake, or additional information may be required.