Final answer:
The limiting reactant is O2, and upon complete reaction of 9.50 g of O2 with CO, 26.14 grams of CO2 could be produced according to the stoichiometry of the balanced chemical equation.
Step-by-step explanation:
The question asks how many grams of CO2 could be produced when 9.50 grams of CO react with 9.50 grams of O2 according to the balanced chemical equation:
2 CO(g) + O2(g) → 2 CO2(g).
First, we need to find the limiting reactant, which is the reactant that will be completely used up first and limit the amount of product formed. To do this, we can calculate the moles of each reactant:
- Moles of CO = Mass / Molar mass = 9.50 g / 28.01 g/mol = 0.339 moles
- Moles of O2 = Mass / Molar mass = 9.50 g / 32.00 g/mol = 0.297 moles
According to the balanced equation, 2 moles of CO react with 1 mole of O2, meaning we need twice as many moles of CO than O2. Since we have more than twice the amount of moles of CO compared to O2, O2 is the limiting reactant. Therefore, all of the O2 will react, and CO will be left in excess.
Now, to find the amount of CO2 produced, we can use the ratio from the balanced equation, which shows 2 moles of CO2 formed for every 1 mole of O2 reacted:
Moles of CO2 produced = Moles of O2 reacted x (2 moles CO2 / 1 mole O2) = 0.297 moles x 2 = 0.594 moles.
To convert moles of CO2 to grams, we use the molar mass of CO2 (44.01 g/mol):
Grams of CO2 produced = Moles x Molar mass = 0.594 moles x 44.01 g/mol = 26.14 g.
Thus, 26.14 grams of CO2 could be produced.