For complete combustion of an organic compound needs 8 molecules of oxygen, what could the formula of this compound be?
1 answer:
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Answer: a) . Ammonia is the limiting reagent
b. Oxygen is left over and 0.1375 g of oxygen is left over.
c. The theoretical yield of is 35.29 g.
d. The theoretical yield of is 31.74 g.
Explanation:
To calculate the number of moles, we use the equation:
For
Given mass of ammonia = 20.0 g
Molar mass of ammonia = 17.031 g/mol
Putting values in equation 1, we get:
For
Given mass of oxygen gas = 50.0 g
Molar mass of oxygen gas = 32 g/mol
Putting values in equation 1, we get:
The chemical equation for the reaction is
By Stoichiometry of the reaction:
4 moles of ammonia reacts with = 5 moles of oxygen
So 1.17 moles of ammonia will react with = of oxygen
As, given amount of oxygen is more than the required amount. So, it is considered as an excess reagent and (1.6-1.4625)= 0.1375 g of oxygen is left unreacted.
Thus ammonia is considered as a limiting reagent because it limits the formation of product.
1. By Stoichiometry of the reaction:
4 moles of ammonia produces = 4 moles of
1.17 moles of ammonia will produce = of
Mass of
Thus Theoretical yield of is 35.29 grams.
2. By Stoichiometry of the reaction:
4 moles of ammonia produces = 6 moles of
1.2 moles of ammonia will produce = of
Mass of
Thus Theoretical yield of is 31.74 grams.