A reaction vessel contains 7.92 g of CO and 7.92 of O2. How many grams of CO2 could be produced according to the follow unbalanced reaction?

Question

CO + O2 --> CO2

Answer 1

2CO + O2 --> 2CO2
This is a limiting reagent (LR) problem
mols CO initially = grams/molar mass = 7.92/28 = approx 0.3
mols O2 initially = 7.92/32 = approx 0.25

Now convert mols CO and mols O2 to mols CO2 that could be produced. This is a LR problem; therefore, you may not produce more than the LEAST amount possible.That is, you could produce 0.3 mol CO2 (0.3 mols CO2 x (2 mol CO2/2 mol CO) if you had 0.3 mol CO initially and all of the O2 needed or 0.25 mol O2 x (2 mol CO2/1 mol O2) = appox 0.5 mol if you had 0.25 mol CO and all of the O2 needed.In LR problem the smaller number is always the correct one; i.e., use the 0.3 mols CO, then convert to grams. grmas = mols x molar mass.Post your work you get stuck. Remember that the 0.3 and 0.2 are approximations and you should do the math more accurately.

Answer 2

A simple way to determine the limiting reactant is to convert reactant quantities into moles and then divide by the respective coefficient of each reactant of the balanced equation. The smallest value is the limiting reactant. Solve for yields based on moles of limiting reactant.

Given 2CO + O₂ => 2CO₂
Moles CO = 7.92g/28g/mol = 0.29 mol CO
Moles O₂ = 7.92g/32g/mol = 0.25 mol O₂
Divide moles of each reactant by respective coefficient of balanced equation…
Moles CO / 2 = 0.29/2 = 0.14
Moles O₂ / 1 = 0.25/1 = 0.25
Smallest number is Limiting Reactant => Limiting Reactant is CO.
Work problem from 0.29 mole CO => 0.29 mole CO₂(44g/mol) = 12.76 g CO₂ produced (theoretical)