Post Module Assessment

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Post Module Assessment

Postby 504986173 » Mon Oct 01, 2018 8:23 pm

I am having trouble on number 19 of the "balancing chemical equations" post module assessment. I'm not sure how to determine the answer; can someone walk me through it?
The question is as follows:

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

Im not sure how they got the products

Justin Haggard 1E
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Joined: Fri Sep 28, 2018 12:18 am

Re: Post Module Assessment

Postby Justin Haggard 1E » Mon Oct 01, 2018 8:37 pm

Combustion of 4 mols of C4H10(g):

Step 1: Start by writing the skeletal equation, or the "bare bones." Combustion uses oxygen to create carbon dioxide and water vapor. This is the unbalanced equation:

C4H10(g) + O2(g) --> CO2(g) + H2O(g)

Step 2: Then note that the problem specified to use 4 mols of C4H10(g), and then balance the carbon atoms in the products.

4C4H10(g) + O2(g) --> 16CO2(g) + H2O(g) **NOTE: This is still unbalanced**

Step 3: Now balance the hydrogen atoms in the products (there are a total of 40 in the reactants, so there needs to be 40 in the products):

4C4H10(g) + O2(g) --> 16CO2(g) + 20H2O(g)

Step 4: Use the balanced products to determine how many oxygen molecules you need in the reactants (there are a total of 52 oxygen atoms in the products, but since oxygen gas molecules consist of a pair of oxygen atoms bonded together, you divide this number by 2):

4C4H10(g) + 26O2(g) --> 16CO2(g) + 20H2O(g)

And there's your balanced equation.

Step 5: Then... to determine the net number of moles produced, compare the moles of reactant to that of the products:

4 moles + 26 moles --> 16 moles + 20 moles

4+26=30 and 16+20= 36, so the net number of moles produced is products-reactants or 36-30= 6

Hope this helps!!

Vy Lu 2B
Posts: 59
Joined: Fri Sep 28, 2018 12:24 am

Re: Post Module Assessment

Postby Vy Lu 2B » Mon Oct 01, 2018 8:44 pm

You would first balance out the original chemical equation C4H10(g) + O2(g) → CO2(g) + H2O(g)
Since the problem calls for 4 moles of C4H10 (g), we add the coefficient to that molecule: 4C4H10(g) + O2(g) → CO2(g) + H2O(g)
After that we simply balance out the rest of the chemical equation following that: since we have 16 C atoms on the left side, we'd add the coefficient 16 to the right side in front of CO2: 4C4H10(g) + O2(g) → 16CO2(g) + H2O(g)
Since there are 40 H atoms on the left side, we'd add the coefficient 20 on the left in front of the H2O: 4C4H10(g) + O2(g) → 16CO2(g) + 20H2O(g)
The last element to balance is oxygen and we do so by adding the coefficient 26 to the O2 on the left side of the equation; now you have your balanced equation: 4C4H10(g) + 26O2(g) → 16CO2(g) + 20H2O(g).

To find the net amount of moles produced, subtract the total amount of moles of product with the total amount of moles of reactant: 36 moles - 30 moles = 6 moles produced in total.

Posts: 31
Joined: Fri Sep 28, 2018 12:29 am

Re: Post Module Assessment

Postby jguiman4H » Mon Oct 01, 2018 8:49 pm

So, we are given that there are 4 moles of butane, C4H10, as well as the chemical equation with varying stoichiometric coefficients.
What we know so far, thus, is: 4 C4H10(g) + ___O2(g) --> ___CO2(g) + ___H20(g)

Typically, we can balance chemical equations using the CHO technique (Carbon-Hydrogen-Oxygen, in this order).
Because we have 4 mol of C4H10(g), we have 16 of C. Therefore, because there is only one C on the products side, we need 16 as our coefficient.
So now, we have: 4 C4H10(g) + ___O2(g) --> 16 CO2(g) + ___H20(g)

Next, looking at hydrogen, there are 10 in the molecule and we still have 4 mol of H, so 4 x 10 gives us 40 H. Then, because we have H20, we would only need 20 of these water molecules because H2 x 20 = 40 H.
So now, we have: 4 C4H10(g) + ___O2(g) --> 16 CO2(g) + 20 H20(g)

In equations like this, we save oxygen for last because it is usually the easiest to balance at the end. We have an unknown amount on the reactants side, but we know that we have 32 O from the CO2 and 20 O from the H20 on the products side. Thus, we have 52 total O. However, because oxygen exists as a diatomic molecule (O2) on the reactants side, we will divide 52 by 2 to determine that we need 26 as our oxygen coefficient.
So now, we have: 4 C4H10(g) + 26 O2(g) --> 16 CO2(g) + 20 H20(g)

Given that this is our final equation, we can now count the moles of both the reactants and products to determine the net number of moles of gas produced. So, we have 4 mol of butane and 26 mol of oxygen on the reactants side. And, we have 16 mol of carbon dioxide and 20 mol of water on the products side. To calculate net number of moles, we will use (mol products - mol reactants), so (16 + 20) - (4 + 26) gives us 36 - 30, so our net mole amount is 6 mol.

This means our final answer will be A.

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