anonymous
  • anonymous
How many grams of magnesium oxide can be produced when 97.2 g Mg react with 88.5 g O2? I balanced the equation as 2Mg + O2 --> 2MgO but I have now gotten two different answers
Chemistry
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
@Rushwr
anonymous
  • anonymous
So I took the 97.2 g Mg and divided by the molar mass of 24.3050 g/mol and got 3.999 mol Mg But since the ratio is 1:1 I don't understand what to do now
Rushwr
  • Rushwr
Hey! This is also easy. 1st find the moles of each compound separately. O2 moles = 88.5g divided by 32gmol^-1 = 2.766mol Mg moles = 97.2g divided by 24.3050gmol^-1 = 3.999mol According to that Mg to O2 ratio is 2:1 right? but looking at the no. of moles of each we can say that the ratio isn't correct right? But we know the reaction will occur in that ratio anyhow ! According to the equation O2 moles needed to react with 3.99 moles of Mg is 1.995 moles. But we have 2.766 moles right? That means we have ecess oxygen. Also means Mg is the limiting factor of the reaction. Amount of Mg limits the reaction. So we have to use the no. of moles of Mg cuz it's the limiting factor. Okai ? As we can see in the equation the ratio of Mg used to MgO produced is 2:2 so it will have the same no. of moles. Hope you can do the rest of it alone ! Did u get it ? :)

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Rushwr
  • Rushwr
Did u get it ?
anonymous
  • anonymous
I think I did I kinda got a bit lost
Rushwr
  • Rushwr
Where did u get it confused ?
anonymous
  • anonymous
How do I use the second ratio like I understand why I have to use them but how?
Rushwr
  • Rushwr
It's like this 1st we have to identify the limiting agent. 2nd ratio luk at the equation. The ratio of both moles of Mg used up to MgO formed is 1:1 right ?
Rushwr
  • Rushwr
that means if u use x moles of Mg x moles of MgO will be formed !
anonymous
  • anonymous
yes and so I multiply 3.999 mol x 3.999 mol?
anonymous
  • anonymous
To get 15.992?
Rushwr
  • Rushwr
no no u don't have to multiply so we just get that idea that the moles of Mg used is equal to the no. of moles of MgO formed right? Since we have asked to workout the grams of MgO use n=m/M n-no. of moles m-mass M-molar mass So we know the no. of moles and molar mass, WE have to find the mass. Substitute the values u will get the mass
Rushwr
  • Rushwr
did u get it ?
anonymous
  • anonymous
So 1.383 + 3.999 g 5.382 / 40.3044 g/mol
anonymous
  • anonymous
I got .1335
Rushwr
  • Rushwr
why did u add ? u have to multiply the no of moles and the molar mass
anonymous
  • anonymous
to get the number of moles the O2 i divided 2.766 by 2 and got 1.383 and added that to Mg 3.999 I am assuming i am wrong
Rushwr
  • Rushwr
you are confused !
Rushwr
  • Rushwr
it's like this Since we are given with the masses of both the reactants we have to find the moles of each reactant. We find it to see if there's any limiting agent. Ex: think we are finding the moles of Mg and O2 after that the ratio of moles were 2:1 exactly there's no limiting agent. But here we found that we don't get that ratio right?
anonymous
  • anonymous
yes oh wait so I do it for each one?
Rushwr
  • Rushwr
for each reactants separately ! We found that earlier In my 1st comment that is given !
anonymous
  • anonymous
for the n= m/M
Rushwr
  • Rushwr
yeah !
anonymous
  • anonymous
ohhhhh thats my fault okay let me try this again
anonymous
  • anonymous
3.999/24.3050 = .1645 2.766/31.9988 = 11.5686 like this?
Rushwr
  • Rushwr
Hey I think U need more help with this!
anonymous
  • anonymous
Its okay if you can't explain it to me ill get it eventually
Rushwr
  • Rushwr
reply me ! I messaged u !

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