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MakcuM [25]
3 years ago
12

A chemist needs to find out how large of a container would be required to contain the products of an ammonia decomposition. If 8

.91 mol H2 are produced during the reaction, how many moles of N2 are produced?
Chemistry
2 answers:
goldenfox [79]3 years ago
8 0
The answer is 2.97 mol N<span>2</span>
Zielflug [23.3K]3 years ago
3 0

Answer : The moles of N_2 produced are, 2.97 moles

Solution : Given,

Moles of H_2 = 8.91 moles

The balanced decomposition reaction of ammonia will be,

NH_3\rightarrow 3H_2+N_2

From the balanced reaction, we conclude that

When 2 moles of NH_3 decomposes to produce 3 moles of H_2 and 1 mole of N_2

If the moles of H_2 is 3 moles then the moles of N_2 produces = 1 mole

If the moles of H_2 is 8.91 moles then the moles of N_2 produces = \frac{8.91moles}{3moles}=2.97moles

Therefore, the moles of N_2 produced are, 2.97 moles

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2 years ago
Question 3) A 1.00 L buffer solution is 0.250 M in HF and 0.250 M in LiF. Calculate the pH of the solution after the addition of
Masja [62]

The pH of the solution after adding 0.150 moles of solid LiF is 3.84

<u>Explanation:</u>

We have the chemical equation,

HF (aq)+NaOH(aq)->NaF(aq)+H2O

To find how many moles have been used in this

c= n/V=> n= c.V

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Simillarly

nF=0.250 M⋅1.5 L=0.375 moles F

nHF=0.375 moles - 0.250 moles=0.125 moles

nF=0.375 moles+0.250 moles=0.625 moles

[HF]=0.125 moles/1.5 L=0.0834 M

[F−]=0.625 moles/1.5 L=0.4167 M

To determine the problem using the Henderson - Hasselbalch equation

pH=pKa+log ([conjugate base/[weak acid])

Find the value of Ka

pKa=−log(Ka)

pH=−log(Ka) +log([F−]/[HF]

pH= -log(3.5 x 10 ^4)+log(0.4167 M/0.0834 M)

pH=-log(3.5 x 10 ^4)+log(4.996)

pH= -4.54+0.698

pH=-(-3.84)

pH=3.84

The pH of the solution after adding 0.150 moles of solid LiF is 3.84

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