Question

4000 grams of heptane (C7H16) is combusted with 7000 grams of oxygen.C7H16 + 11 O2 —> 7 CO2 + 8 H2OWhat is the limiting reactant?How many grams of carbon dioxide is produced?

Answer 1

To find the limiting reactant, we have to convert the given masses to moles of each substance, using their molecular masses:

$4000g\cdot\frac{mol}{100g}=40mol$$7000g\cdot\frac{mol}{31g}=225.8mol$

Use the mole ratio to determine the amount of one of the reactants to the other one:

$40\text{molC}7H16\cdot\frac{11\text{ molO2}}{1\text{molC7H16}}=440molO2$

There is not enough oxygen to react with the given amount of heptane. It means that the limiting reactant is oxygen.

Use the mole ratio to find the amount of carbon dioxide produced:

$225.8molO_2\cdot\frac{7molCO_2}{11molO_2}=143.7molCO_2$

Convert the answer to grams using the molecular mass of carbon dioxide:

$143.7molCO_2\cdot\frac{44g}{mol}=6322.8g$

The answer is 6322.8 grams.