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ICE Princess25 [194]
3 years ago
10

How many liters of ammonia gas can be formed from 54.1 liters of hydrogen

Chemistry
1 answer:
Aleks04 [339]3 years ago
5 0
 <span>The reaction is N2 + 3H2 -> 2NH3 
So the amount of NH3 formed is 2/3 of the amount of H2 = 2/3 * 13.7 = 9.13 Liters.</span><span>The answer is 9.13

</span>
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2 years ago
AlBr3 can be used as a catalyst in the Friedel-Crafts alkylation reaction. The correct name for the compound represented by the
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The question is incomplete, the complete question is;

AlBr3 can be used as a catalyst in the Friedel-Crafts alkylation reaction. The correct name for the compound represented by the formula AlBr3 is —

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aluminum bromide

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Having known that AlBr3 is an ionic compound and aluminium is the central atom here, we now have to ask ourselves if Aluminium exists in other stable oxidation states.

We must take cognizance of the fact that the oxidation number of the central atom in a compound becomes part of the name of that compound when other stable oxidation states for atoms of the same elements exists.

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7 0
2 years ago
Why can't methanol, CH3OH, be used as a solvent for sodium amide, NaNH2? Sodium amide is nonpolar and methanol is polar. Sodium
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sodium amide is a ionic compound and basically exists as sodium cation and amide anion. Amide anion is highly basic in nature and hence as soon as  there is amide anion generated in the solution , Due to its very pronounced acidity it very quickly abstracts the slightly acidic proton available on methanol.

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Hence sodium amide reacts with methanol and abstracts its only acidic proton and form ammonia and sodium Methoxide.

Hence the 3rd statement is a corrects statement.

So we cannot use methanol for sodium amide because sodium amide itself would react with methanol and the inherent molecular natur of sodium amide would then change.

The 1st and 2nd statements both are incorrect because both the compounds methanol as well as sodium amide have dipole moments and hence are polar molecules.

The 4th statement is also incorrect as both the molecules have dipole moment and hence there would be ion-dipole forces operating between them.

The following reaction occurs:

NaNH₂+CH₃OH→NH₃+CH₃ONa

4 0
3 years ago
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