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uranmaximum [27]
3 years ago
9

The limiting reactant is the chemical substance that determines the amount of product(s) that can ultimately be formed in a reac

tion. During the reaction, the limiting reactant is completely consumed or used up, and therefore, causes the reaction to stop. The limiting reactant can be identified through stoichiometric calculations. After comparing the results, the reactant that produces the smaller mass of product is identified as the limiting reactant. Determine the limiting reactant and calculate the number of grams of Al2O3 that can be formed when 10.0 grams of Al and 5.00 grams of O2 react. 4 Al(s) + 3 O2(g) → 2 Al2O3(s) For the calculations in this module, the molar mass of an element will be rounded to the hundredths place (0.01 g).
Chemistry
1 answer:
Mariulka [41]3 years ago
6 0

Explanation:

Al=10,o2=5

(Al10)2(o5)3

The answer is oxygen.

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1.)Which of the following is an appropriate action if you are caught outside in a thunderstorm?
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4 years ago
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Your lab partner combined chloroform and acetone to create a solution where the mole fraction of chloroform, Xchloroform, is 0.1
jeyben [28]

Answer:

Explanation:

[u]Assumptions[/u]

1. There is exactly 1 mole of chloroform

2. The liquids mix together well such that the volume of the solution is the sum of the volumes of the two liquids

Given that the mole fraction of the Chloroform is 0.171

Mole fraction of Chloroform =

Mole of chloroform /(Mole of chloroform + mole of acetone)

According to assumptions, mole of chloroform is equal to 1

Therefore 0.171 =1/(1+mole of acetone)

1 + mole of acetone = 1/0.171

Mole of acetone = 1(/0.171) - 1

Moles of acetone = 4.85mol.

From Stochiometry

Mass of acetone = Mole of acetone * Molar Mass of acetone

Molar mass of acetone = 58.1grams/mol

Mass of acetone = 4.85 *58.1 = 282g =0.282kg

Mass of chloroform = moles of chloroform *Molar mass of Chloroform

Molar mass of chloform = 119.4 grams/mol

Mass of chloroform = 1* 119.4 =119.4g=0.1994kg

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Volume of acetone = 357mL

Volume of Chloroform = Mass of Chloroform /Density of Chloroform

Volume of Chloroform = 119.4/1.48

= 81mL

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Molarity = 1/0.282 = 3.55molal

2. Molarity = moles of solute( chloroform) /Volume of solution

= 1/0.438 =2.28Molar

Therefore the molality and molarity respectively are 3.55 and 2.28.

4 0
4 years ago
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I believe the answer is B??????????? Hope this helps

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