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

1 Answer
Dr_P
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_2H_3OF = 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_2H_3OF 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_2H_3OF x 1 mol H_2O /1 mol C_2H_3OF = 0.675 mol H_2O.

This is the maximum amount of water that would form if ALL C_2H_3OF is consumed.

1.91 mol O_2 x 1 mol H_2O /1 mol O_2 = 1.91 mol H_2O .

This is the maximum amount of water that would form if ALL O_2 is consumed.

Since complete use of all the C_2H_3OF will produce the least (0.675 mol H_2O of water, then C_2H_3OF is the limiting reactant.

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