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Chemistry 10 Online
OpenStudy (anonymous):

Methane, CH4, burns in oxygen gas to form water and carbon dioxide. What is the correct balanced chemical equation for this reaction? I'm having a hard time understanding how to balance a chemical equation. I'm really not sure how to even begin. Please help!

OpenStudy (anonymous):

also, another question. How many moles of Helium are in 16.0 g of the element? I put 16 mol but got it wrong, it was 4 mol. I don't understand...

OpenStudy (anonymous):

because I thought the molar mass is the same as the mass in grams..

OpenStudy (accessdenied):

For your first question, have you written out the skeleton equation for the reaction first? If that term is unfamiliar, I mean the equation without any coefficients put in yet; just the molecules interacting. That's a great starting point.

OpenStudy (anonymous):

you mean CH4+02 ---->H2O+CO2 ?

OpenStudy (accessdenied):

Yep. That equation is useful as a starting point. To balance the equation, we basically want to insert coefficients on each atom in order to make the number of atoms of each element the same on both sides. We'd start with the first element and work our way through them all. So here we can start with carbons: \( \large \rm \color{red}CH_4 + O_2 \to H_2 O + \color{red}CO_2 \) We see that there is one carbon atom on the left and one on the right. So the carbons look good for now. Next we can look at the hydrogen atoms. \( \large \rm C\color{red}{H_4} + O_2 \to \color{red}{H_2} O + CO_2 \) Here we have four hydrogen atoms on the left and two on the right. So we need to insert a coefficient that multiplies with 2 to get 4. That'd be 2. \( \large \rm C\color{red}{H_4} + O_2 \to \color{maroon}2 \color{red}{H_2} O + CO_2 \) Then we just have oxygen last. Do you think you see how this process works and can figure out what to do with the oxygen atoms?

OpenStudy (anonymous):

I know it'd be CH4+2O2 ---> 2H2O+CO2 but why would the coefficient in front of O2 on the left be 2 if it's just O on the right? @AccessDenied

OpenStudy (accessdenied):

Well, there is one O in the H2O atom. The coefficient of 2 doubles the number of each atom, so there would be two O's total now. Then we add in the two from CO2 as well, making a total of 4 on the right.

OpenStudy (anonymous):

ugh I'm so confused

OpenStudy (accessdenied):

\( \large \rm CH_4 + \color{red}{O_2} \to \color{red}2 H_2 \color{red}{O} + C\color{red}{O_2} \) Does it help to view it like this? Highlighting the oxygen atoms.

OpenStudy (anonymous):

ohhh I wasn't counting the oxygen in 2H2O as being 2.. the coefficient isn't just for the hydrogen it's for the oxygen too?

OpenStudy (accessdenied):

Yep. A coefficient is kind of like a coefficient in math. \( 2x = x + x \), and \( \rm2 H_2 O = H_2 O + H_2 O\), or literally two \(\rm H_2 O\) molecules.

OpenStudy (anonymous):

ok. oh my gosh I understand now. thank you so much

OpenStudy (accessdenied):

Glad to help! As for this question: "How many moles of Helium are in 16.0 g of the element?" For this, you would look up the molar mass of Helium, which is about 4 g/mol. You could look at this in the sense of a ratio problem. \( \dfrac{4 \ g}{1 \ mol} = \dfrac{16 \ g}{x \ mol} \). This could be solved by Algebra, or also notice that from 4 g to 16 g is a multiple of 4, so the 1 mol to x mol is a multiple of 4 as well. 1 x 4 = 4 mol

OpenStudy (accessdenied):

Or alternatively, the way I always do it is by unit conversion. You have the grams and you want moles. The molar mass is basically a conversion factor between the two (the unit is grams per mole). So starting with grams, we need to multiply by moles per gram, or divide by grams per mole. In math writing (so its easier to read): \( \text{grams} \times \dfrac{\text{moles}}{\text{grams}} = \dfrac{\text{grams}}{\text{grams/mole}} = \dfrac{\text{mas}s}{\text{molar mass}}\) So you have 16 g, and the molar mass is 4 g/mol; thus 16 g / (4 g/mol) = 16/4 mol = 4 mol.

OpenStudy (anonymous):

but I'm not allowed to look stuff up when I'm taking a quiz. how would I know the molar mass??

OpenStudy (accessdenied):

This information would be available on a periodic table. If I'm not mistaken those should usually be provided for stuff like quizzes!

OpenStudy (anonymous):

oh wait because molar mass is the same as the atomic mass right? and the atomic mass is under the symbol omg wow duh

OpenStudy (accessdenied):

Yup, they're the same! If you didn't have a periodic table for a quiz, I'd hope they'd at least use the most common elements in that case of molar masses. H, C, N, and O (and occasionally Cl) come to mind first from the chemistry I've done so far.

OpenStudy (anonymous):

you are a life saver. I'm taking the act on Saturday so this will help immensely. thank you so much

OpenStudy (accessdenied):

Glad to help out! :)

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