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Alenkasestr [34]
3 years ago
13

What volume of ammonia in liters, at STP, can be produced from 2.0 kg of hydrogen gas, H2, and an excess of nitrogen gas, N2? N2

(g) + 3 H2 (g) → 2 NH3 (g)
Chemistry
1 answer:
Sonbull [250]3 years ago
4 0
Ok so rewrite the equation for yourself :).

N2 + 3H2 -> 2NH3 (Always check if it is balanced).

Ok so we have 2kg of hydrogen conver this to grams and then to moles.
2kg -> 2000g
2000g divide by R.M.M of H2
R.M.M
2H - 2 * 1 -> 2
2000/2 - 1000 moles.
Look at the ratio of H2 to 2NH3.
We see that we get 2NH3 for every H2 therefore the ratio is 1:2.
This means that we HAVE 2000 moles of 2NH3.
At stp. we know that 1 moles = 22.4 Litres
We have 2000 moles that means that we produce 44800 litres of NH3 at stp.
2000 * 22.4 - 44800
Hope this helps :).

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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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