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3 years ago

What is the mass in grams of H₂ that can be formed from 54.6 grams of NH₃ in the following reaction?

Chemistry

1 answer:

kaheart [24]3 years ago

5 0

Answer : The mass of $H_2$ is, 9.64 grams.

Explanation : Given,

Mass of $NH_3$ = 54.6 g

Molar mass of $NH_3$ = 17 g/mol

Molar mass of $H_2$ = 2 g/mol

First we have to calculate the moles of $NH_3$ .

$\text{Moles of }NH_3=\frac{\text{Given mass }NH_3}{\text{Molar mass }NH_3}$

$\text{Moles of }NH_3=\frac{54.6g}{17g/mol}=3.21mol$

Now we have to calculate the moles of $H_2$

The balanced chemical equation is:

$2NH_3(g)\rightarrow 3H_2(g)+N_2(g)$

From the balanced reaction we conclude that

As, 2 mole of $NH_3$ react to give 3 moles of $H_2$

So, 3.21 mole of $NH_3$ react to give $\frac{3}{2}\times 3.21=4.82$ mole of $H_2$

Now we have to calculate the mass of $H_2$

$\text{ Mass of }H_2=\text{ Moles of }H_2\times \text{ Molar mass of }H_2$

$\text{ Mass of }H_2=(4.82moles)\times (2g/mole)=9.64g$

Therefore, the mass of $H_2$ is, 9.64 grams.

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Predict the splitting pattern for each of the labeled hydrogens in the following molecules. Assume that all coupling constants a

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Complete Question

The complete question is shown on the first uploaded image

Answer:

a) Splitting pattern for Ha= 2+1 , Triplet

For Proton Hb and Hc both are equivalent to each other. non equivalent protons n= 3

Splitting pattern for Hb and Hc= 3+1 , Quartet

For Proton Hd, number of non equivalent protons n= 0

Splitting pattern for Hd= 0+1 , Singlet

b) Splitting pattern for Ha= 2+1 , Triplet

For Proton Hb and Hc both are equivalent to each other. non equivalent protons n= 3

Splitting pattern for Hb and Hc= 3+1 , Quartet

For Proton Hd, number of non equivalent protons n= 0

Splitting pattern for Hd= 0+1 , Singlet

c) The IUPAC name is Butan-2-ol

Explanation:

Considering the first question the rule used for prediction of splitting pattern is n+1 (Pascal's Triangle rule), where n is number of H atom on the adjacent carbon which are non equivalent.

According to that for molecule 1 as shown on the second uploaded image

For Proton Ha, number of non equivalent protons n= 2

Splitting pattern for Ha= 2+1 , Triplet

For Proton Hb and Hc both are equivalent to each other. non equivalent protons n= 3

Splitting pattern for Hb and Hc= 3+1 , Quartet

For Proton Hd, number of non equivalent protons n= 0

Splitting pattern for Hd= 0+1 , Singlet

Considering the second question for Molecule 2 as shown on the third uploaded image

For Proton Ha, number of non equivalent protons n= 1

Splitting pattern for Ha= 1+1=2 , Doublet

For Proton Hb, number of non equivalent protons n= 3

Splitting pattern for Hb= 3+1=4 , Quartet

For Proton Hc, number of non equivalent protons n= 3

Splitting pattern for Hc= 3+1=4 , Quartet

For Proton Hd, number of non equivalent protons n= 1

Splitting pattern for Ha= 1+1=2 , Doublet

Considering the third question

The name of the given molecule is gotten according to longest carbon chain = 4 (Prefix 'Butan')

Functional group = -OH (Suffix 'ol') at C-2

The IUPAC name is Butan-2-ol

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Answer:

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