Question 8 on UA's Practice Test

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Alexander Hari 1L
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Joined: Wed Feb 14, 2018 3:02 am

Question 8 on UA's Practice Test

Postby Alexander Hari 1L » Wed Apr 11, 2018 5:45 pm

"For the following reaction, how many grams of AlCl3 would be obtained if 5.43g of aluminum and 7.80g of hydrogen chloride (HCl) was used in the reaction?
Al (s) + HCl (aq) → AlCl3 (aq) + H2 (g)"

I've figured out that the limiting reactant in the equation is HCl and that we have 0.21 moles of HCl. But what do I do now to figure out how many grams of AlCl4 were produced? Do I multiply by the molar mass of AlCl3? Thanks

Andrew Evans - 1G
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Re: Question 8 on UA's Practice Test

Postby Andrew Evans - 1G » Wed Apr 11, 2018 6:00 pm

Once you know the limiting reactant to be HCl, you just have to use the moles of HCl and convert it to moles of AlCl3. Since you know the molar ratio from the balanced equation (6 HCl : 2 AlCl3), you use that to convert from moles HCl to moles AlCl3. And then, yes, multiply the moles of AlCl3 produced by the molar mass to get the amount of grams AlCl produced.

–Andrew Evans
Section 1G

jessicasam
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Re: Question 8 on UA's Practice Test

Postby jessicasam » Wed Apr 11, 2018 8:44 pm

Now that you know that you have .21 moles HCl, you want to get ?gAlCl3

.21moles HCl (1 mole AlCl3/ 1 mole HCl) (133.34 g AlCl3/mole AlCl3)= 28 g AlCl3

Nicole Shak 1L
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Joined: Wed Nov 22, 2017 3:03 am

Re: Question 8 on UA's Practice Test

Postby Nicole Shak 1L » Fri Apr 13, 2018 11:26 am

Another way to do this problem using dimensional analysis is to start with both grams given and convert to g of ALCL3, so this way you can determine the limiting reactant and g of ALCL3 all in one step.


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