## Page F109 from book (Self-Test M.2A)

William Lan 2l
Posts: 73
Joined: Fri Sep 29, 2017 7:07 am

### Page F109 from book (Self-Test M.2A)

How do you identify the limiting reactant in the reaction 6 Na + Al2O3 ->2 Al + 3 Na2O when 5.52 g of sodium is heated with 5.1 g of Al2O3?

Lena Nguyen 2H
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Joined: Fri Sep 29, 2017 7:06 am
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### Re: Page F109 from book (Self-Test M.2A)

To identify the limiting reactant, you need to use the amounts of each reactant in moles.

So converting each reactant to moles,
(5.52g Na)(1 mol/22.99g Na) = .24 mol Na
(5.1g Al2O3)(1 mol/101.957g Al2O3) = .05 mol Al2O3

For every 1 mol of Al2O3, you need 6 mol of Na to complete the reaction because of the stoichiometric coefficients. So if you want to use up all .05 moles of Al2O3, you would need (.05 x 6 = .3) .3 moles of Na. Since there is only .24 mol Na, Na is in shortage and is therefore the limiting reactant.

I hope this helps!

Mika Sonnleitner 1A
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Joined: Fri Sep 29, 2017 7:04 am
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### Re: Page F109 from book (Self-Test M.2A)

To find the limiting reactant, you would first convert the given masses into moles using the molar masses for each reactant. Then, using the balanced equation, you can see that 1 mole of Al2O3 requires 6 moles of Na. When you calculate the moles of Al2O3 and multiply by 6 (to find the moles of Al2O3 you need), the moles of Na is less than the amount you require, therefore Na is the limiting reactant. In summary:

1) Convert 5.52g Na into moles by using the molar mass of Na (22.99 g/mol), then convert 5.10 g Al2O3 into moles by using its molar mass (101.96 g/mol)
2) Multiply the moles of Al2O3 by 6 (to calculate the moles of Al2O3 you need)
3) If the moles of Al2O3 you calculated in step 1 is less than the moles of Al2O3 you calculated in step 2, then Na is the limiting reactant, since you need more Na than you have.

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