A reaction was performed in which 0.55 g of 2‑naphthol was reacted with a slight excess of allyl bromide to make 0.59 g of allyl
1 answer:
Answer:
- Theoretical yield = 0.7028 g
- Percent yield = 83.95%
Explanation:
The reaction that takes places is:
- C₁₀H₈O + C₃H₅Br → C₁₃H₁₂O + HBr
The<u> theoretical yield</u> is calculated when assuming that <em>all of the limiting reactant transformed into the product</em>:
- 0.55 g naphthol ÷ 144gNaphthol/mol *
* 184 gEther/mol = 0.7028 g
The <u>percent yield</u> is calculated using the theoretical yield and the experimental yield:
- 0.59g / 0.7028 g * 100% = 83.95%
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Answer:
Explanation:
Given parameters:
Mass of aluminium oxide = 3.87g
Mass of water = 5.67g
Unknown:
Limiting reactant = ?
Solution:
The limiting reactant is the reactant in short supply in a chemical reaction. We need to first write the chemical equation and convert the masses given to the number of moles.
Using the number of moles, we can ascertain the limiting reactants;
Al₂O₃ + 3H₂O → 2Al(OH)₃
Number of moles;
Number of moles =
molar mass of Al₂O₃ = (2x27) + 3(16) = 102g/mole
number of moles = = 0.04mole
molar mass of H₂O = 2(1) + 16 = 18g/mole
number of moles = = 0.32mole
From the reaction equation;
1 mole of Al₂O₃ reacted with 3 moles of H₂O
0.04 mole of Al₂O₃ will react with 3 x 0.04 mole = 0.12 mole of H₂O
But we were given 0.32 mole of H₂O and this is in excess of amount required.
This shows that Al₂O₃ is the limiting reactant
Answer:
a. alkyne
b. alkane
c. alkyne
d. alkene
Explanation:
The general formula for each class of compound is given below
Alkane:
Alkene:
Alkyne: (assuming single multiple bonds)
Now let us classify according to the above formulas:
a. It has two hydrogen atoms less than the two times of carbon atoms hence, it's alkyne
b. It has two hydrogen atoms more than the two times of carbon atoms hence, it's alkane
c. It has two hydrogen atoms less than the two times of carbon atoms hence, it's alkyne
d. It has hydrogen atoms two times of carbon atoms hence, it's alkene