Question

Q1. 3000 grams of heptane is combusted with 25000 grams of oxygen. C7H16 + O2 --> CO2 + H2O a) What is the limiting reactant? b) How many grams of carbon dioxide is produced? c) How many grams of excess reactant are left? *

Answer 1

(a)

Heptane is the limiting factor. as O2 can be found in air.

(b)

balanced equation: C7H16 + 11 O2 --> 7 CO2 + 8 H2O

moles of O2= mass/ mr

moles of O2= 25000/32

moles of O2 = 781.25 moles

According to molar ratio:

11 : 7

moles of carbon dioxide:

(781.25/11) * 7

497.16 moles

using mass = moles * mr

mass of CO2 = 497.16 * 44 = 21875.04 g -- this is mass of CO2