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3 years ago

6.5 Moles of Al reacts with 7.2 Moles of H2O What is the limiting reactant and calculate the Theoretical Yield?

Chemistry

1 answer:

SVEN [57.7K]3 years ago

4 0

Answer:

H₂O is the limiting reactant

Theoretical yield of 240 g Al₂O₃ and 14 g H₂

Explanation:

Find how many moles of one reactant is needed to completely react with the other.

6.5 mol Al × (3 mol H₂O / 2 mol Al) = 9.75 mol H₂O

We need 9.75 mol of H₂O to completely react with 6.5 mol of Al. But we only have 7.2 mol of H₂O. Therefore, H₂O is the limiting reactant.

Now find the theoretical yield:

7.2 mol H₂O × (1 mol Al₂O₃ / 3 mol H₂O) × (102 g Al₂O₃ / mol Al₂O₃) ≈ 240 g Al₂O₃

7.2 mol H₂O × (3 mol H₂ / 3 mol H₂O) × (2 g H₂ / mol H₂) ≈ 14 g H₂

Since the data was given to two significant figures, we must round our answer to two significant figures as well.

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sergeinik [125]

Answer:

Electrones de valencia.

Explanation:

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3 years ago

Click the "draw structure" button to launch the drawing utility. under certain reaction conditions, 2,3−dibromobutane reacts wit

REY [17]

Answer:

Explanation:

According to this. Let's analize the possible products a, b and c.

First, the problem states that we have 2 eq. of base, For this case, let's assume it's KOH. Now, As we are doing a reaction with base, means that this reaction can only take places under conditions of SN2 and E2, a fast reaction that is taking place in only 1 step.

With this in mind, let's analyze product a. This states that it has two sp hybridized carbon, in other words, a triple bond between two carbons. So the product is with no doubt, an alkyne.

Product b has only one sp hybridized carbon, which means that this carbon should cannot be an alkyne because we need two carbon atoms. The only way to have one atom of C sp hybridized, is with two double bonds, so product b would have to a alkene with two double bonds.

Product c do not have sp hybridized carbon, therefore, it only has two double bonds in two different Carbon atoms, so it's another alkene with two double bonds, but in two different atoms of carbon.

Picture attached show the product a, b and c. Hope this can help

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3 years ago

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3 years ago

What is the mass of HF produced by three reaction of 3.0 10 to the 23 molecules of H2 with excess F2

JulijaS [17]

Answer:

It is 20. g HF

Explanation:

H2 + F2 ==> 2HF ... balanced equation

Since the question is asking us to find the mass of product formed, we will want to first convert the molecules of H2 into moles of H2 (we could do this at the end of the calculations, but it's just as easy to do it now).

moles of H2 present (using Avogadro's number):

3.0x1023 molecules H2 x 1 mole H2/6.02x1023 molecules = 0.498 moles H2

From the balanced equation, we see that 1 mole H2 produces 2 moles HF. Therefore, we can now find the theoretical mass of HF produced from 0.498 moles H2:

0.498 moles H2 x 2 moles HF/1 mol H2 = 0.996 moles HF formed.

The molar mass of HF = 20.01 g/mole, thus...

0.996 moles HF x 20.01 g/mole = 19.93 g HF = 20. g HF formed (to 2 significant figures)

6 0

3 years ago

Consider the nuclear equation below. 239/94 Pu—-> X+ 4/2 He. What is X?

Anna007 [38]

Answer:

X = U (Uranium)

Explanation:

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4 0

3 years ago

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