Question

2. C3Hs 5 O2 3CO2 +4 H20 Initial Grams Initial moles Change in moles Final Moles Final Grams Fill in the amounts for the table above given that 100g of C3Hs and 170 g of oxygen are initially available to react. Round to three decimal places when necessary

Answer 1

C3H8 + 5O2 -----------> 3CO2 + 4H2O

Initial mass of C3H8 = 100 grams

Initial Moles of C3H8 = mass/molar mass = 100/44 = 2.273 moles

Initial mass of O2 = 170 grams

Intial moles of O2 = mass/molar mass = 170/32 = 5.313 moles

Initial moles and initial mass of H2O and CO2 will be zero because reaction has not been started yet.

From reaction, it is clear that 1 mole of C3H8 requires 5 moles of O2.

So, 2.273 moles of C3H8 will requires 5*2.273 = 11.365 moles of O2.

However available moles of O2 = 5.313 which is less than the required one i.e. 11.365.

Again from reaction,

5 moles of O2 require 1 mole of C3H8

So, 5.313 moles of O2 will require 5.313/5 = 1.0626 moles of C3H8

So, O2 is limiting reagent and it will be used completely by C3H8 and C3H8 is excess reagent. So, some part of C3H8 will remain un-reacted.

Final available moles of O2 = initial - used = 5.313 - 5.313 = 0 moles

Final available moles of C3H8 = initial - used = 2.273 - 1.0626 = 1.2104 moles

Final mass of O2 = moles*molar mass = 0*32 = 0 grams

Final mass of C3H8 = 1.2104*44 = 53.258 grams

Again from reaction, 5 mole O2 produce 3 mole CO2

So, 5.313 moles O2 will produce 3*5.313/5 = 3.188 moles of CO2

So, final moles of CO2 = 3.188 moles

Final mass of CO2 = 3.188*44 = 140.272 grams.

Again from reaction, 5 moles of O2 produce 4 moles of H2O

So, 5.313 moles O2 will produce 4*5.313/5 = 4.2504 moles of H2O

So, final moles of H2O = 4.2504

Final mass of H2O = 4.2504*18 = 76.507 grams