Question

Suppose you buy 40.46 pounds of propane for your gas grill.
b. According to the following reaction for the combustion of propane, what is the mass (in grams) of water that will be produced by the combustion of the entire tank? C₃H₈(l) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(l)

___ g H₂O

c. What mass of oxygen is required for this combustion?

___ g O₂

d. If 0.1599 moles of propane and 0.6243 moles of oxygen are sealed in a container and burned, which reactant will be the limiting reactant, and how many moles of carbon dioxide will be produced?
Limiting reactant is [oxygen]

___ mol CO₂ produced

Answer 1

b) Weight of propane = 40.46 pounds = 40.46 ^ 453.6 g =18352.66 g.   [ as 1 pound = 453.6 gram ]

From Stoichiometry of Given reaction: C₃H₈(l) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(l) we can say that,

44 grams of propane (C₃H8) produces 72 grams H₂O (18^4=72)

So, 18355.66 g of propane will produce (72/44)^18355.66 g H₂O = 30036.53 g H₂O

So, 30036.53 g H₂O will be produced by entire tank.

c) Also, 44 g of propane reacts with 80 g of O₂ (16^5=80)

So, 18352.66 g of propane will react with (80/44)^18352.66 = 33368.47 g of O₂

So, 33368.47 g of O₂ is required fro this combustion.

d) From the stoichiometry of above reaction it can be seen that 1 mole of propane produces 3 moles of CO₂

So, From 0.1599 moles of propane, no. of moles of CO₂ produced = 0.1599^3 = 0.4797 moles of CO₂

Also, 5 moles of O₂ produces 3 moles of CO₂

So, 0.6243 moles of oxygen will produce (3/5)^0.6243 moles of CO₂ = 0.37458 moles of CO₂

As Oxygen produces less no. of moles of CO_2, It means oxygen will be the limiting reagent. And it will produce 0.37458 moles of CO_2.

Please give it a thumbs-up if you like my answer. Feel free to ask your queries in the comment box. Thank you.