1 meso-stilbene dibromide + 2 KOH ——–> 1 Diphenylacetylene + 2 H2O + 2 KBr The reaction is run on 0.0015 mole of meso-stilbene dibromide. Calculate how many grams are needed. A 50% molar excess of KOH over the stoichiometrically required amount is used. KOH pellets (~85% KOH content) typically weigh about 0.1 g and are not easily divided. Because a considerable excess is used, the precise mass is not critical. Determine the number of pellets required to give roughly the calculated number of moles. To begin the calculation, find the stoichiometrically required number of moles based on the balanced equation. Calculate the expected yield of the product, diphenylacetylene, based on the limiting reactant.
The Correct Answer and Explanation is:
To solve this problem, we need to calculate several steps based on stoichiometry. Let’s break it down.
Step 1: Find the stoichiometrically required moles of KOH
The reaction is as follows:meso-stilbene dibromide+2 KOH→Diphenylacetylene+2 H2O+2 KBr\text{meso-stilbene dibromide} + 2 \, \text{KOH} \rightarrow \text{Diphenylacetylene} + 2 \, \text{H}_2\text{O} + 2 \, \text{KBr}meso-stilbene dibromide+2KOH→Diphenylacetylene+2H2O+2KBr
From the equation, we can see that 1 mole of meso-stilbene dibromide reacts with 2 moles of KOH.
You are given 0.0015 moles of meso-stilbene dibromide, so:Required moles of KOH=2×0.0015=0.003 moles of KOH\text{Required moles of KOH} = 2 \times 0.0015 = 0.003 \, \text{moles of KOH}Required moles of KOH=2×0.0015=0.003moles of KOH
Step 2: Find the moles of KOH needed with a 50% excess
Since a 50% molar excess of KOH is used, we need to add half again to the stoichiometrically required moles:Excess KOH=0.003 moles×1.5=0.0045 moles of KOH\text{Excess KOH} = 0.003 \, \text{moles} \times 1.5 = 0.0045 \, \text{moles of KOH}Excess KOH=0.003moles×1.5=0.0045moles of KOH
So, you will need 0.0045 moles of KOH for the reaction.
Step 3: Find the mass of KOH required
Now, we need to calculate the mass of KOH that corresponds to 0.0045 moles. The molar mass of KOH is approximately 56.11 g/mol. Thus:Mass of KOH=0.0045 mol×56.11 g/mol=0.2525 g\text{Mass of KOH} = 0.0045 \, \text{mol} \times 56.11 \, \text{g/mol} = 0.2525 \, \text{g}Mass of KOH=0.0045mol×56.11g/mol=0.2525g
Step 4: Account for the purity of KOH (85% KOH content)
The KOH pellets are 85% pure. Therefore, the actual mass of KOH pellets required to give 0.2525 grams of pure KOH will be:Mass of KOH pellets=0.2525 g0.85=0.2965 g\text{Mass of KOH pellets} = \frac{0.2525 \, \text{g}}{0.85} = 0.2965 \, \text{g}Mass of KOH pellets=0.850.2525g=0.2965g
Step 5: Find the number of KOH pellets
Each pellet weighs about 0.1 g. To find how many pellets are needed:Number of pellets=0.2965 g0.1 g/pellet=2.97≈3 pellets\text{Number of pellets} = \frac{0.2965 \, \text{g}}{0.1 \, \text{g/pellet}} = 2.97 \approx 3 \, \text{pellets}Number of pellets=0.1g/pellet0.2965g=2.97≈3pellets
Step 6: Calculate the yield of diphenylacetylene
Since the limiting reactant is meso-stilbene dibromide (and we are using a sufficient excess of KOH), the theoretical yield of diphenylacetylene will be equal to the number of moles of meso-stilbene dibromide used.
From the reaction, 1 mole of meso-stilbene dibromide produces 1 mole of diphenylacetylene. Therefore, the yield will be:Moles of diphenylacetylene=0.0015 moles\text{Moles of diphenylacetylene} = 0.0015 \, \text{moles}Moles of diphenylacetylene=0.0015moles
Final Answer:
- Mass of KOH pellets needed: Approximately 0.2965 g (or 3 pellets).
- Theoretical yield of diphenylacetylene: 0.0015 moles (or 0.0015 moles × the molar mass of diphenylacetylene if needed for mass).
