How many moles are in 68.16 grams of NH3 ?
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Percentage yield shows the amount of reactants converted into products. The percentage yield of the reaction is 51.7%.
The equation of the reaction is sown in the image attached. The reaction is 1:1 as we can see.
Number of moles of NaBr = 10 g/103 g/mol = 0.097 moles
We can obtain the mass of 3-methyl-1-butanol from its density.
Mass = density × volume
Density of 3-methyl-1-butanol = 0.810 g/mL
Volume of 3-methyl-1-butanol = 9 mL
Mass of 3-methyl-1-butanol = 0.810 g/mL × 9 mL
Mass of 3-methyl-1-butanol = 7.29 g
Number of moles of 3-methyl-1-butanol = mass/molar mass = 7.29 g/88 g/mol = 0.083 moles
Since the reaction is 1:1 then the limiting reagent is 3-methyl-1-butanol
Mass of product 1-bromo-3-methylbutane = number of moles × molar mass
Molar mass of 1-bromo-3-methylbutane = 151 g/mol
Mass of product 1-bromo-3-methylbutane = 0.083 moles × 151 g/mol
= 12.53 g
Recall that % yield = actual yield/theoretical yield × 100
Actual yield of product = 6.48 g
Theoretical yield = 12.53 g
% yield = 6.48 g/12.53 g × 100
% yield = 51.7%
Learn more: brainly.com/question/5325004
