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Answer :
If 1.70g of aniline reacts with the 2.10 g of bromine, the theoretical yield of 4-bromoaniline is 2.25 g.
1 mole mole of the aniline react with the 1 mol of Brâ‚‚ produces 1 mole of the 4 - bromoaniline.
mass of aniline = 1.70 g
molar mass of aniline = 93 g/mol
moles of aniline = 0.0182 mol
mass of Brâ‚‚ = 2.10 g
molar mass of Brâ‚‚ = 160 g/mol
moles = 2.10/ 160
= 0.0131 mol
Brâ‚‚ is limiting reagent .
moles of 4-bromoaniline = 0.0131
mass of 4-bromoaniline = moles × molar mass
= 0.0131 × 172
= 2.25 g
Thus, the theoretical yield is 2.25 g.
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Final answer:
The question involves the reaction of aniline with bromine to form 4-bromoaniline, which is a chemical reaction in the field of chemistry. By identifying the limiting reagent, in this case bromine, and using stoichiometry based on the balanced chemical equation, we can find the theoretical yield of 4-bromoaniline, which is about 2.06 grams.
Explanation:
This question involves understanding the stoichiometry of the reaction between aniline and bromine to form 4-bromoaniline. The balanced chemical equation for this reaction is:
C6H7N + Br2 -> C6H6BrN + HBr. Assuming the reaction goes to completion, the limiting reagent is the one of which we have fewer moles, since it will be completely consumed first and thus dictate the maximum yield of product. To determine this, we need to convert the masses of aniline and bromine into moles. 1.70 g of aniline (with a molar mass of 93.13 g/mol) is 0.0182 moles, and 2.10 g of bromine (with a molar mass of 159.808 g/mol) is 0.0131 moles. Bromine is therefore the limiting reagent. According to the balanced equation, one mole of bromine produces one mole of 4-bromoaniline. Hence, the theoretical yield of 4-bromoaniline is 0.0131 moles. Converting this to grams using the molar mass of 4-bromoaniline (157.02 g/mol), we find the theoretical yield to be approximately 2.06 g.
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