# Answer to Question #57109 in General Chemistry for tasha

Question #57109
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.)
1
2015-12-22T12:32:22-0500
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.

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