# Stoichiometry homework help

## Homework Statement

Given the balanced equation 2Al(NO3)3 + 3Mg ==> 3Mg(NO3)2 + 2Al , if 54.1 g of aluminum nitrate is reacted, what mass of magnesium nitrate is produced?

2. The attempt at a solution
I've been given an entire packet dedicated to stoichiometry, but nothing in here covers this problem. I am unsure of how to even begin solving it. Maybe if I had help with the first few steps I could solve the rest.

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a good way to go about this problem is to convert the mass of aluminum nitrate into moles of aluminum nitrate, then you can figure out how many moles of magnesium (do you mean aluminum or magnesium nitrate) were produced.

Ok so I converted the grams of aluminum nitrate into moles and got approx. 0.2539 moles. I'm still confused about what to do next, though.

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Ok so I converted the grams of aluminum nitrate into moles and got approx. 0.2539 moles. I'm still confused about what to do next, though.

Now you need to use the mole fraction, which you get from the balanced reaction.

What is the mole fraction? Like, 2 moles of aluminum nitrate to 3 moles of magnesium nitrate?

What is the mole fraction? Like, 2 moles of aluminum nitrate to 3 moles of magnesium nitrate?

Yes.
(2mol Al nitrate/3mol Mg nitrate)

Ok, so how do I use this ratio in relation to the 0.2539 moles of aluminum nitrate? I guess I still don't really get all of the steps.

Ok, so how do I use this ratio in relation to the 0.2539 moles of aluminum nitrate? I guess I still don't really get all of the steps.

It is a dimensional analysis setup

mol Al nitrate(mole fraction)(convert mol Mg nitrate to g)=soln

So I take 0.2539 and multiply it by 2/3 first, right?

Never mind, I got the right answer. Thanks for the help. :)