Balancing Equation Post Module Assessment

Moderators: Chem_Mod, Chem_Admin

Mahnoor_Wani_1I
Posts: 124
Joined: Wed Sep 30, 2020 9:35 pm
Been upvoted: 3 times

Balancing Equation Post Module Assessment

Postby Mahnoor_Wani_1I » Fri Oct 23, 2020 8:19 pm

19. During a summer camping weekend 4 moles of butane (C4H10) gas were used for cooking. Chose the right balanced equation for the combustion of 4 moles of butane gas. What is the net number of moles of gas produced?

A. 4C4H10(g) + 26O2(g) → 16CO2(g) + 20H2O(g); 6

B. 4C4H10(g) + 26O2(g) → 16CO2(g) + 18H2O(g); 4

C. 4C4H10(g) + 25O2(g) → 16CO2(g) + 20H2O(g); 6

D. 4C4H10(g) + 26O2(g) → 16CO2(g) + 20H2O(g); 5

I was able to find the chemical equation to be : 4C4H10(g) + 26O2(g) → 16CO2(g) + 20H2O(g) but I don't know. how to find the net number of moles. Isn't just moles of CO2 + H20?

Alvin Lai 3J
Posts: 101
Joined: Wed Sep 30, 2020 9:55 pm

Re: Balancing Equation Post Module Assessment

Postby Alvin Lai 3J » Fri Oct 23, 2020 8:22 pm

So the coefficients in front of each compound is a mole. All you have to do is add up the total moles on the left and compare them to the total moles on the right. This should be 30->36, so the net moles is 6.

Carly_Lipschitz_3H
Posts: 104
Joined: Wed Sep 30, 2020 9:56 pm

Re: Balancing Equation Post Module Assessment

Postby Carly_Lipschitz_3H » Sun Oct 25, 2020 3:37 pm

Basically it is just asking about the total moles on the reactants side vs the product side. I added up all the moles on each side. Then, I subtracted the number of moles on the products side from the reactants side to see the net number of moles of gas produced. So, it ends up being 36-30 = 6. Net moles is just asking to to see how many moles were gained or lost during the reaction.


Return to “Balancing Chemical Reactions”

Who is online

Users browsing this forum: No registered users and 2 guests