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MA_775_DIABLO [31]
4 years ago
11

Draw structures for the carbonyl electrophile and enolate nucleophile that react to give the enone below. You do not have to con

sider stereochemistry. Draw the enolate ion in its carbanion form. Draw one structure per sketcher. Add additional sketchers using the drop-down menu in the bottom right corner. Separate multiple reactants using the sign from the drop-down menu.

Chemistry
1 answer:
sweet-ann [11.9K]4 years ago
7 0

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

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During the chemical reaction given below 21.71 grams of each reagent were allowed to react. Determine how many grams of the exce
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Answer: 16.32 g of O_2 as excess reagent are left.

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} SO_2=\frac{21.71g}{64g/mol}=0.34mol

\text{Moles of} O_2=\frac{21.71g}{32g/mol}=0.68mol

2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)  

According to stoichiometry :

2 moles of SO_2 require = 1 mole of O_2

Thus 0.34 moles of SO_2 will require=\frac{1}{2}\times 0.34=0.17moles  of O_2

Thus SO_2 is the limiting reagent as it limits the formation of product and O_2 is the excess reagent.

Moles of O_2 left = (0.68-0.17) mol = 0.51 mol

Mass of O_2=moles\times {\text {Molar mass}}=0.51moles\times 32g/mol=16.32g

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