Calculate the mass of magnesium oxide formed when 10.0 grams of Mg burns in excess oxygen

- anonymous

Calculate the mass of magnesium oxide formed when 10.0 grams of Mg burns in excess oxygen

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- anonymous

do you have a balanced equation?

- anonymous

@jcab98

- anonymous

Yep
Its 2Mg(s)+O2(g)=2MgO(s)

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## More answers

- anonymous

If oxygen is excess then magnesium is the limiting reagent. I like the dimensional analysis method because you can pretty much do everything in 1 step.|dw:1436102167827:dw||dw:1436102167827:dw|

- anonymous

idk why that did that twice

- anonymous

if you notice, all the units cancel except g MgO and all you have to do is multiply and divide the numbers
|dw:1436102693078:dw|

- anonymous

@peachpi could you describe the method you used for the above answer because this method isnew to me

- anonymous

Dimensional analysis is where you treat the parts of a chemical equation as units that you can convert from one form to another using the ratios in the equation. Once the equation is balanced all the compounds/elements are in a fixed ratio, so what you're essentially doing is scaling up or down depending on what you start with or want to end with.
^that seems kind of like a word salad :/
It's kind of like converting miles to cm. You could convert miles to feet first, then in another step feet to inches, then in another inches to cm. Or you could set it all up once as a series of ratios to be multiplied
Does that make sense?

- anonymous

@peachpi so what did you do to start off.Idont understand how you got the numbers on the top and the bottom

- anonymous

I debated whether putting them in vs having you fill them in...especially since we might be using different periodic tables.
The 24.31is the molar mass of Mg. Basically molar mass is in units of g/mol. As in there are 24.31 g of Mg in 1 mole of Mg. The grams go on bottom because you want it to cancel with the 10 g given in the problem

- anonymous

so that leaves the 1 mol Mg to go on top

- anonymous

with me so far?

- anonymous

yes @peachpi

- anonymous

ok. the 2's come from the coefficients of the balanced equation.
After the first step we have mol Mg on top. We need to convert that to mol MgO, That's why 2 mol MgO goes on top and 2 mol Mg is on bottom. This is hands down the most important part where you're going from one chemical to another. If the equation isn't balanced it won't be right. This one works out to a 1:1 ratio, but that won't always be the case

- anonymous

Continue @peachpi im starting to understand now

- anonymous

ok. you now have mol MgO. Since this problem wants grams you have to use the molar mass for MgO to change moles to grams. I got 40.31 g/mol for the molar mass, so 40.31 g MgO goes on top and 1 mol MgO goes on bottom.
That's the reverse of the way we did it at the beginning of the problem when we were going from grams to moles. This time we're doing moles to grams

- anonymous

g MgO is what is asked for in the problem so that's the last conversion.

- anonymous

do the top part of the working out represent what you want to find out ?

- anonymous

yes it does. and the bottom is what you have/ need to cancel

- anonymous

when you mean need to cancel is that with the10.0g Mg

- anonymous

FYI, See
http://www.chembuddy.com/?left=balancing-stoichiometry&right=dimensional-analysis
and
http://www2.sunysuffolk.edu/sambass/index_files/studyguide/SA-5.pdf
for more info on the method. The first link has another example, coincidentally using the same formula. The 2nd has a summary of the method in general

- anonymous

or the previous step

- anonymous

every step will cancel. So yes the first step is canceling the 10 g with g

- anonymous

Is this method only used for this type of question

- anonymous

how do you mean?

- anonymous

you can use it in any application where you're converting elements with a constant ratio, if that's what you mean

- anonymous

For only calcualting the mass of products when given a mass of a reactant

- anonymous

What other applications are there

- anonymous

no you can actually use it for mole to mole, moles to gram, etc. Those are actually a little shorter than this one.
If you had 10 mole of Mg and wanted to find g MgO for example, you wouldn't need to do that first conversion with the molar mass. You'd start with 10 mol Mg then jump straight to the ratios from the equation.
Similar thing if you had moles Mg and wanted moles MgO. That middle conversion would be the only step.

- anonymous

As for other applications, I used it a lot in physics like when finding speeds given rotation or converting between english/metric units for mass and force

- anonymous

Thank you for your help @peachpi .For my first question on this site the information you provided me will really help me in my chemistry studies.
Btw What country are you from and what grade are you in

- anonymous

you're welcome! happy to help. I'm from the Bahamas and I've been out of school a loooong time. I tutor so I just come on to keep my brain fresh.

- anonymous

Well thank you for your help and your time.I'm currrently in Year 11 Studying Chemistry for the first time in Australia .Not doing that badly at it ,coming 2nd in the grade.I'm just revising over calculations for chemistry, because its a hard concept to grasp

- anonymous

You're welcome. Good luck to you. It's a fascinating subject.

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