Suppose much of an old forest will be used for farmland. A certain portion of the land will be set aside as wildlife habitat. Yo
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Answer:
1/8
Explanation:
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Answer:
81.3 %
Explanation:
The formula for the calculation of moles is shown below:
For ferrocene:-
Mass of ferrocene = 100 mg = 0.1 g
Molar mass of ferrocene = 186.04 g/mol
The formula for the calculation of moles is shown below:
Thus,
For acetyl chloride:-
Volume = 40 microliters = 0.04 mL
Density = 1.1 g / mL
Density is defined as:-
or,
Mass of acetyl chloride = 0.044 g
Molar mass of acetyl chloride = 78.49 g/mol
The formula for the calculation of moles is shown below:
Thus,
As per the reaction stoichiometry, one mole of ferrocene reacts with one mole of acetyl chloride to give one mole of monoacetylferrocene
Limiting reagent is the one which is present in small amount. Thus, ferrocene is limiting reagent.
The formation of the product is governed by the limiting reagent. So,
one mole of ferrocene on reaction forms one mole of monoacetylferrocene
0.0005375 mole of ferrocene on reaction forms 0.0005375 mole of monoacetylferrocene
Moles of product formed = 0.0005375 moles
Molar mass of monoacetylferrocene = 228.07 g/mole
Mass of monoacetylferrocene produced = Moles*molecular weight = 0.0005375*228.07 g = 0.123 grams = 123 mg
Given experimental yield = 100 mg
<u>% yield = (Experimental yield / Theoretical yield) × 100 = (100/ 123) × 100 = 81.3 %</u>