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denis23 [38]
3 years ago
6

When 6.0 mol Al react with 13 mol HCl, what is the limiting reactant, and how many moles of H2 can be formed? 2Al + 6HCl → 2AlCl

3 + 3H2
Chemistry
1 answer:
4vir4ik [10]3 years ago
6 0

Well, we need to find the ratio of Al to the other reactant.


Al:HCl = 1:3


--> this means that for every 1 Al used, you have to use 3 HCl.



6*3 = 18 moles of HCl needed to fully react with 6 moles of Al. Since 13<18, HCL is the limiting reactant.



The ratio of HCl:AlCl = 3:1



13/3 = 4.3333...



The final answer is HCl is the limiting reactant with 4.3 moles of AlCl3 able to be produced.



Hope this helps!!! :)


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Part A. Two containers, one at 305 K and the other at 295 K, are placed in contact with each other. 1. 1 J of heat flows from th
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0.00011 JK.

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Explanation:

The following parameters are given which are going to help in solving for the change in entropy of the system. The term "entropy'' simply means the degree of disorderliness of a system.

=> The temperature of container A = 305 K, the temperature of container B = 295 K and the amount of heat generated when the containers are placed in contact with each other = 1. 1 J.

The change in entropy of the hot container = -(1/305) = - 0.00328 J/K.

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Step 2: Calculate moles

Moles = mass / molar mass

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Moles C = 42.88 grams / 12.01 g/mol

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Moles O = 57.12 grams / 16.0 g/mol

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Step 3: Calculate the mol ratio

We divide by the smallest amount of moles

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O: 3.57 moles / 3.57 moles = 1

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