In a reaction where an OH group is substituted with a Cl, 15.02 g of 3-phenyl-2-butanol (MW=150.2 g/mol) is the starting materia
1 answer:
Answer:
16.81 g of C₁₀H₁₃Cl
Explanation:
When 3-phenyl-2-butanol reacts with HCl it undergoes a substitution reaction to form 2-chloro-3-phenylbutane and water.
C₁₀H₁₄O + HCl ⇄ C₁₀H₁₃Cl + H₂O
The molar mass of 3-phenyl-2-butanol is 150.2 g/mol and the molar mass of 2-chloro-3-phenylbutane is 168.1 g/mol. The molar ratio is 1:1. Then, for 15.02 g of 3-phenyl-2-butanol, the theoretical yield is:
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Answer:
Explanation:
In this case, we can start with the reaction:
If we check the reaction, we will have 2 X and Y atoms on both sides. So, <u>the reaction is balanced</u>. Now, the problem give to us two amounts of reagents. Therefore, we have to find the <u>limiting reagent</u>. The first step then is to find the moles of each compound using the <u>molar mass</u>:
Now, we can <u>divide by the coefficient</u> of each compound (given by the balanced reaction):
The smallest value is for "X", therefore this is our <u>limiting reagent</u>. Now, if we use the <u>molar ratio</u> between "X" and "XY" we can calculate the moles of XY, so:
Finally, with the molar mass of "XY" we can calculate the grams. Now, we know that 1 mol X = 85 g X and 1 mol = 48 g
(therefore 1 mol Y = 24 g Y). With this in mind the <u>molar mass of XY</u> would be 85+24 = 109 g/mol. With this in mind:
I hope it helps!