Answer:
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Answer:
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Answer:
%
Explanation:
The ethanol combustion reaction is:
→
If we had the amount (x moles) of ethanol, we would calculate the oxygen moles required:

Dividing the previous equation by x:

We would need 3.30 oxygen moles per ethanol mole.
Then we apply the composition relation between O2 and N2 in the feed air:

Then calculate the oxygen moles number leaving the reactor, considering that 0.85 ethanol moles react and the stoichiometry of the reaction:

Calculate the number of moles of CO2 and water considering the same:


The total number of moles at the reactor output would be:

So, the oxygen mole fraction would be:
%
Answer:
12 grams of hydrogen gas
and 56 grams of nitrogen gas
The molar mass of ammonia is 17 g/mol.
68 grams of ammonia corresponds to
17g/mol
68g
=4moles
4 moles of ammonia will be obtained from
2
4×1
=2 moles of nitrogen and
2
4×3
=6 moles of hydrogen.
The molar masses of nitrogen and hydrogen are 28 g/mol and 2 g/mol respectively.
2 moles of nitrogen corresponds to 2×28=56 grams.
6 moles of hydrogen corresponds to 6×2=12 grams.