14)
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<u>Answer:</u> The percent yield of the reaction is 91.8 %
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
.....(1)
- <u>For
:</u>
Given mass of = 4.0 g
Molar mass of = 63.12 g/mol
Putting values in equation 1, we get:
- <u>For oxygen gas:</u>
Given mass of oxygen gas = 10.0 g
Molar mass of oxygen gas = 32 g/mol
Putting values in equation 1, we get:
The chemical equation for the reaction of and oxygen gas follows:
By Stoichiometry of the reaction:
12 moles of oxygen gas reacts with 2 moles of
So, 0.3125 moles of oxygen gas will react with = of
As, given amount of is more than the required amount. So, it is considered as an excess reagent.
Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
12 moles of oxygen gas produces 5 moles of
So, 0.3125 moles of oxygen gas will produce = of water
Now, calculating the mass of from equation 1, we get:
Molar mass of = 69.93 g/mol
Moles of = 0.130 moles
Putting values in equation 1, we get:
To calculate the percentage yield of , we use the equation:
Experimental yield of = 8.32 g
Theoretical yield of = 9.052 g
Putting values in above equation, we get:
Hence, the percent yield of the reaction is 91.8 %