Im stuck on these two problems
1 answer:
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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
Answer:
empirical formula:
2 g H
32.7 g S
65.3 g O
Explanation:
Like the problem said, the first thing we can do is calculate the mass of each of the 3 elements in a 100-gram sample:
- 2.00% * 100g = 2 g H
- 32.7% * 100g = 32.7 g S
- 65.3% * 100g = 65.3 g O
Now we need to find the empirical formula from these. To do so, convert all of those masses into moles by using the molar mass for each element:
- the molar mass of H is 1.01 g/mol
- the molar mass of S is 32.06 g/mol
- the molar mass of O is 16 g/mol
2 g H ÷ 1.01 g/mol = 1.98 mol H
32.7 g S ÷ 32.06 g/mol = 1.02 mol S
65.3 g O ÷ 16 g/mol = 4.08 mol O
Our ratio of H : S : O is now:
1.98 mol : 1.02 mol : 4.08 mol
Divide them all by the smallest number, which is 1.02:
1.98/1.02 : 1.02/1.02 : 4.08/1.02
1.94 : 1 : 4
1.94 ≈ 2
So:
2 : 1 : 4
Thus, the empirical formula is: .