Question 4.13: Limiting Reagent . Suppose 12 g of C is mixed with 64 g of O...

(a) We use the molar mass of each reactant to calculate the number of moles of each compound present before reaction.

12 \cancel{g C} \times \frac{1 mol C }{12 \cancel{g C}} =1 molC

64 \cancel{g O_{2} } \times \frac{1 mol O_{2} }{32 \cancel{g O_{2}}} =2 mol O_{2}

According to the balanced equation, reaction of one mole of C requires
one mole of O_{2}. But two moles of O_{2} are present at the start of the reaction. Therefore, C is the limiting reagent and O_{2} is in excess.

(b) To calculate the number of grams of CO_{2} formed, we use the conversion factor 1 mol CO2 = 44 g CO2.

12 \cancel{ g C}\times \frac{1 \cancel{mol C}}{12 \cancel{ g C}}\times \frac{1 \cancel{mol CO_{2} }}{1 \cancel{mol C} } \times \frac{44g CO_{2}}{1 \cancel{mol CO_{2} }} = 44 g CO_{2}

We can summarize these numbers in the following table. Note that, as required by the law of conservation of mass, the sum of the masses of the material present after reaction is the same as the amount present before any reaction took place, namely 76 g of material.