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vampirchik [111]
3 years ago
15

You want to recrystallize the product from a nucleophilic substitution reaction. After adding a minimal amount of hot 95% ethano

l to dissolve all the crystals and letting the flask cool to room temperature crystals do not form. To induce crystallization and maximize the % recovery, one should:_______.
A) Boil off some solvent
B) Scratch the flask
C) Add a seed crystal
D) Cool the flask in an ice bath
E) Any of the above
Chemistry
1 answer:
Troyanec [42]3 years ago
6 0

Answer:

B) Scratch the flask

C) Add a seed crystal

Explanation:

Two important methods to induce crystallization in this case are;

1) Scratch the flask

2) Add a seed crystal

However, the rate of crystallization becomes slower at lower temperature hence cooling in an ice bath should only be applied before the crystals form. When the crystals start forming, the system should be maintained at room temperature.

The question clearly stated that a minimal amount of hot 95% ethanol was added to dissolve all the crystals hence you do not need to boil off some solvent. Boiling off solvent is applied when too much solvent is used.

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Answer:

a) Limiting: sulfur. Excess: aluminium.

b) 1.56g Al₂S₃.

c) 0.72g Al

Explanation:

Hello,

In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.

b) By stoichiometry, the produced grams of aluminium sulfide are:

m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3

c) The leftover is computed as follows:

m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.

Best regards.

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