Question

Consider the following reaction:
2CO(g)+O2(g)→2CO2(g). a reaction mixture contains 28 g of CO and 32 g of O2, what is the limiting reactant? (Try to do this problem in your head without any written calculations.)

Answer 1

As we see from the reaction:
2CO(g)+O₂(g)→2CO₂(g);
Every 2 molecules of CO react with 1 molecule of O₂. Therefore, at the equilibrium point, molar proportion of CO:O₂ is 2:1.
To measure the amount of substance of each gas, we should divide its mass by its molar mass: v=m/M.
M(CO)=28 g/mol
M(O₂)=32 g/mol
Moles of CO: v(CO)=m(CO)/M(CO)=28g / 28g/mol = 1 mol
Moles of CO: v(O₂)=m(O2)/M(O₂)=32g / 32g/mol = 1 mol.
So, v(O₂)=v(CO)=1 mol. But every 1 mol of O₂ will react with two moles of CO, therefore CO is the limiting reagent in our case.