Consider the following unbalanced chemical equation.

Question

C₅H₁₂(l) + O₂(g) → CO₂(g) + H₂O(l)

If 40.0 grams of pentane (C₅H₁₂) are burned in excess oxygen, how many grams of H₂O will be produced?

 

______ g H₂O

Also,

LiOH(s) + CO₂(g) → Li₂CO₃(s) + H₂O(l)

If 87.7 grams of lithium hydroxide reacts with excess carbon dioxide, what mass of lithium carbonate will be produced?

______ g Li₂CO₃

Answer 1

Hello, Jessica,

We need to balance the chemical equation first. Count the numbers of atoms that come in with the reactants, and compare that to those consumed by the products. I like to do this with pencil and paper so that you can make changes easily until you get something that works.

I'll assume you know the basics of balancing. If not, I have a more detailed explanation here: https://drive.google.com/file/d/1CkKeVpdNTTwd95TqKT_taFSCMSX8nE6z/view?usp=sharing

The example is a different equation, but the same principles apply. The important part is to stay organized and start with the most complex molecule first. In this case it is the pentane. You need at least one molecule of each compound in the equation, so pencil in a "1" for the pentane. From that, you see that you need 5 carbons. The only place carbons can go is the CO2, so jot down a "5" in front of it.

The pentane brings in 12 hydrogens. They can only go to the H2O, so put a "6" in front of H2O. Both of the products have oxygen, which can only come from the O2. We already have guessed the numbers of CO2 and H2O, so add the oxygens there, which comes to 16 total. To get 16 oxygens, we need 8 O2 molecules. And after double-checking, we can see the equation is now balanced.

We want to know how many grams of H2O will be produced with 40 grams of pentane. Our equation says that we'll get 6 moles of H2O for every 1 mole of C5H12. So let's calculate the number of moles of pentane.

Moles = mass of pentane/molar mass of pentane. Add the atomic weights of each atom in the molecule. For pentane that is 72.1 AMU. Then simply rewrite the same number as 72.1 grams pentane/mole pentane. Part of the power of the periodic table is the ease for calculating a number that can be used measure out numbers of molecules by their mass, instead of counting them. One mole of anything is equal to Avogadro's number (6.02 x 10²³ ). So if we measure out 72.1 grams of pentane, we'll have that number of pentane molecules. This equation tells us that for that one mole coming in, we'll get 6 moles of H2O, or 3.61 x 10²⁴ molecules of H2O coming out

40 grams of pentane = 40 grams pentane/(72.1 grams pentane/mole pentane). This calculates to 0.555 moles of pentane (3 sig figs). Assuming it all reacts with the excess oxygen, we should get 6 times that number of moles of H2O. Therefore, ((6moles H2O)/(1 mole pentane))*(0.555 moles pentane) = moles of H2o produced. I calculate that to be 3.33 moles H2O.

They want the answer in grams H2O, so convert moles to grams using the molar mass of water (18 g/mole).

3.33 moles H2O * 18 g H2O/mole H2O = 19.9 grams H2O

I hope that helps. Use the same procedure to calculate the second problem. Start with the most complex molecukle when balancing, which I judge to be the Li2CO3. You can see that you need at least 2 Li on the reactant side, if you start with 1 molecule of Li2CO3 on the product side.

Once it is balanced, convert the lithium hydroxide into moles and then use the balanced equation to calculate how many moles of lithium carbonate will be produced, assuming excess CO2.

Have fun,

Bob

Answer 2

1). First balance the equation. Notice there are an even number of O's on the left and an odd number on the right. That means we will need to put an even coefficient in front of the H₂O to get an even number of O's on both sides. You can see that the left side has H₁₂ so putting a 6 in front of the H₂O will solve the "even O's issue" as well as balancing the H's. The CO₂ needs a 5 coefficient because of the C₅ on the left. Now we just need to balance the O's with the proper coefficient in front of the O₂ on the left:

C₅H₁₂(l) + 8O₂(g) → 5CO₂(g) + 6H₂O(l)

Calculate the molar mass of C₅H₁₂ as 5(12.011) + 12(1.008) = 72.151 g/mol. Dividing 40.0 g by 72.151 g/mol gives us 0.55439 moles of C₅H₁₂. In looking at the balanced chemical reaction equation. there are 6 moles of H₂O form every mole of C₅H₁₂ so there would be 6(0.55439) or 3.32636 moles of H₂O produced. Then, using the molar mass of H₂O (which is 1(15.999) + 2(1.008) or 18.015 g/mol), we calculate the mass of H₂O as (3.32636)(18.015) = 59.9 g.

2) Again, balance the equation. That can be achieved by just adding a 2 as the coefficient in front of the LiOH.

2LiOH(s) + CO₂(g) → Li₂CO₃(s) + H₂O(l)

Using the molar mass of LiOH (1•6.941 + 1•15.999 + 1•1.008 = 23.948 g/mol) we can calculate the number of moles as 87.7/23.948 = 3.6621 moles. The balance chemical reaction equation shows there are 2 moles of LiOH required to produce 1 mole of Li₂CO₃ so divide the 3.6621 by 2 to get 1.83105 moles of Li₂CO₃. Using the molar mass of Li₂CO₃ (2•6.941 + 1•12.011 + 3•15.999 or 73.89 g/mol) we can calculate the mass of Li₂CO₃ as (1.83105)(73.89) = 135 g

Answer 3

First, make the balanced equation for the complete combustion of pentane: C₅H₁₂ + 8O₂ ==> 5CO₂ + 6H₂O ...

Then, use dimensional analysis to find mass of H₂O produced:

40 g C₅H₁₂ x 1 mol/72 g x 6 mol H₂O/mol C₅H₁₂ x 18 g H₂O/mol H₂O = 60 g H₂O

Make the balanced equation: 2LiOH(s) + CO₂(g) → Li₂CO₃(s) + H₂O(l)

Then calculate the molar equivalence: Moles of lithium hydroxide=87.7⋅g / 23.95⋅g⋅mol⁻¹=3.66⋅mol.

Given excess carbon dioxide gas, we get a mass of (3.66⋅mol/2)×73.89⋅g⋅mol⁻¹=135.29 g Li₂CO₃