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

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William Lan 2l
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Page F109 from book (Self-Test M.2A)

Postby William Lan 2l » Wed Oct 04, 2017 10:02 pm

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

Postby Lena Nguyen 2H » Wed Oct 04, 2017 10:36 pm

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

Postby Mika Sonnleitner 1A » Wed Oct 04, 2017 10:43 pm

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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