What would be the limiting reagent if 41.9 grams of C2H3OF were reacted with 61.0 grams of O2? C2H3OF+2O2=>2CO2+H2O+HF

May 4, 2014

Always remember to think in terms of the mol in order to solve a problem like this.

First, check to ensure that the equation is balanced (it is). Then, convert the masses to mols: 41.9 g ${C}_{2} {H}_{3} O F$ = 0.675 mol, and 61.0g ${O}_{2}$ = 1.91 mol.

Now, remember that the limiting reactant is the one that restricts how much product forms (ie, it is the reactant that runs out first).

Pick one product, and determine how much would form first if ${C}_{2} {H}_{3} O F$ runs out, AND THEN, if ${O}_{2}$ runs out. To make it easy, if possible, pick a product that has a 1:1 ratio with the reactant you are considering.

0.675 mol ${C}_{2} {H}_{3} O F$ x 1 mol ${H}_{2} O$ /1 mol ${C}_{2} {H}_{3} O F$ = 0.675 mol ${H}_{2} O$.

This is the maximum amount of water that would form if ALL ${C}_{2} {H}_{3} O F$ is consumed.

1.91 mol ${O}_{2}$ x 1 mol ${H}_{2} O$ /1 mol ${O}_{2}$ = 1.91 mol ${H}_{2} O$ .

This is the maximum amount of water that would form if ALL ${O}_{2}$ is consumed.

Since complete use of all the ${C}_{2} {H}_{3} O F$ will produce the least (0.675 mol ${H}_{2} O$ of water, then ${C}_{2} {H}_{3} O F$ is the limiting reactant.