3 0.5 points A student performs the Observation and Classification of Chemical Reactions-Aluminum experiment. They start with 0.034 moles of Aluminum (s) and they conduct the various reactions described in the lab manual, which are given below in their unbalanced form. Assuming no product is lost throughout the experiment, what is the theoretical yield of Alum (s) in grams? The molar mass of Alum is 474.39 g/mol. Round your answer to 2 decimal places. Do not include units in your answer. 1) Al (s) + HCl (aq) →AlCl3 (aq) + H2 (g) 2) AlCl3 (aq) + K2SO4 (aq) + H2O(l) →KAl(SO4)2 ·12 H2O (s) + KCl (aq) Type your answer...

3 0.5 points A student performs the Observation and Classification of Chemical Reactions-Aluminum experiment. They start with 0.034 moles of Aluminum (s) and they conduct the various reactions described in the lab manual, which are given below in their unbalanced form. Assuming no product is lost throughout the experiment, what is the theoretical yield of Alum (s) in grams? The molar mass of Alum is 474.39 g/mol. Round your answer to 2 decimal places. Do not include units in your answer. 1) Al (s) + HCl (aq) \rightarrow AlCl_3 (aq) + H_2 (g) 2) AlCl_3 (aq) + K_2SO_4 (aq) + H_2O(l) \rightarrow KAl(SO_4)_2 \cdot 12 H_2O (s) + KCl (aq) Type your answer...

Transcript

00:01 Okay, so it looks like we're looking at a chemical reaction here, and it's a really complicated one.

00:09 But you don't ever be intimidated by a complicated chemical reaction, especially when we're talking about moles and grams.

00:17 Because when we're talking about moles and grams, we're really only relating two terms to each other or converting using a mass on the periodic table.

00:27 So the first question says they gave us 2 .583 grams of aluminum.

00:33 And they want to know how many moles of aluminum that is.

00:36 Well, remember, when you're looking at the periodic table, one mole of aluminum is going to be equal to 26 .982 grams of aluminum.

00:54 Okay.

00:55 But we don't have 26 .982 grams.

00:58 We only have 2 .583 grams.

01:00 So we want to figure out what that is in moles.

01:04 So we're going to set up a basic dimensional analysis.

01:07 So i'm going to take my 2 .583 grams of aluminum, and i'm going to multiply it by this information that i have over here.

01:19 I'm going to put the one mole of aluminum on the top and the 26 .9.

01:26 982 grams of aluminum on the bottom.

01:31 Now, why did i set that up that way? well, first of all, i need my final unit to be moles of aluminum.

01:39 And i also need my grams of aluminum to cancel.

01:43 And the only way to get those to cancel out would be to have them divide.

01:48 So when i divide the 2 .583 grams by the 26 .9182, grams, i get 0 .0, sorry, 0 .0 .9573.

02:13 And i'm keeping this to three numbers, four numbers, basically because this is four numbers or four significant figures.

02:20 And that would be moles of aluminum.

02:24 So now they want to know how many moles of alum could be produced.

02:30 If we started with that number of aluminum.

02:34 So alum, of course, is the most complicated substance you see in this reaction.

02:42 Now, the thing is, what happens in a chemical reaction is this is like a recipe.

02:49 So two moles of aluminum are going to produce two moles of alum.

02:54 Basically, the word mole can be inserted right there.

02:59 So every two of these will give you two of these.

03:02 But do i have two moles of aluminum? no, i have 0 .09573 moles of aluminum.

03:17 Now remember, i'm doing this based on dimensional analysis.

03:22 So even though this seems kind of tedious, you should really write this step out.

03:26 There's two moles of a of k -a -al s -o -4 -2.

03:36 12h -2.

03:38 This is a hydrate, right, a decahydrate, over two moles of aluminum.

03:45 And i'm writing this out just so i can see that my units are canceling.

03:50 This unit and this unit will be canceled.

03:52 And if i'm multiplying by two and dividing by two, my numerical answer is still going to be the same.

04:00 The only thing that has changed here is my unit is no longer moles of aluminum...

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