In the following reaction, how many liters of oxygen will react with 270 liters of ethene (C2H4) at STP?
1 answer:
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Answer:
The answer to your question is Molarity = 0.41
Explanation:
Data
mass of KNO₃ = 76.6 g
volume = 1.84 l
density = 1.05 g/ml
Process
1.- Calculate the molecular mass of KNO₃
molecular mass = 39 + 14 + (16 x 3) = 101 g
2.- Calculate the number of moles
101 g of KNO₃ --------------- 1 mol
76.6 g of KNO₃ ------------ x
x = (76.6 x 1) / 101
x = 0.76 moles
3.- Calculate molarity
Molarity =
Substitution
Molarity =
Result
Molarity = 0.41
<u>Answer:</u> NO is the limiting reagent in the given reaction and 0.857 moles of will be produced.
<u>Explanation:</u>
Limiting reagent is defined as the reagent which is present in less amount and it limits the formation of products.
Excess reagent is defined as the reagent which is present in large amount.
For the given chemical reaction:
We are given:
Moles of NO = 0.857 mol
Moles of oxygen = 0.498 mol
By stoichiometry of the reaction:
If 2 moles of NO reacts with 1 mole of oxygen gas.
So, 0.857 moles of NO will react with = of
As, the given amount of oxygen gas is more than the required amount. So, it is considered as an excess reagent.
Thus, NO is considered as the limiting reagent because it limits the formation of products.
By Stoichiometry of the reaction:
If 2 moles of NO produces 2 moles of nitrogen dioxide gas.
So, 0.857 moles of NO will produce = of
Hence, NO is the limiting reagent in the given reaction and 0.857 moles of will be produced.