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Nata [24]
3 years ago
9

Of the following reactions occurring at 25ºC, which one involves the greatest increase in entropy?H2(g) + Cl2(g) = 2 HCl(g)H2O(s

) = H2O(l)Pb2+(aq) + 2 Cl-(aq) = PbCl2(s)CO2(s) = CO2(g)
Chemistry
1 answer:
SVETLANKA909090 [29]3 years ago
3 0

Answer: [tex]CO_2(s)\rightarrow CO_2(g)[/tex]

Explanation:

Entropy is the measure of randomness or disorder of a system. If a system moves from an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.

\Delta S is positive when randomness increases and \Delta S is negative when randomness decreases.

a) H_2(g)+Cl_2(g)\rightarrow 2HCl(g)

2 molesof gas are converting to 2 moles of another gas , thus \Delta S is zero.

b) H_2O(s)\rightarrow H_2O(l)

1 mole of solid is converting to 1 mole of liquid, the randomness increases and thus \Delta S is positive.

b) Pb^{2+}(aq)+2Cl^-(aq)\rightarrow PbCl_2(s)

2 moles of ions are converting to 1 mole of solid, the randomness decreases and thus \Delta S is negative

d) CO_2(s)\rightarrow CO_2(g)

1 mole of solid is converting to 1 mole of gas, the randomness increases drastically and thus \Delta S is highly positive.

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sineoko [7]

Answer:

The answer to your question is 1.49 mol of Cl₂  

Explanation:

Data

mass of Al = 40.5 g

mass of Cl₂ = 212.7 g

moles of excess reactant = ?

- Balanced chemical reaction

                2 Al(s)  +  3Cl₂(g)  ⇒   2AlCl₃

Process

a) Calculate the molar mass of the reactants  

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molar mass of Cl₂ = 6 x 35.45 = 212.7 g

b) Calculate the theoretical proportion  Al/Cl₂ = 53.96/212.7 = 0.254

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As the experimental proportion is lower than the theoretical proportion we conclude that the limiting reactant is Aluminum.

c) Calculate the grams of excess reactant

                    53.96 g of Al ------------------ 212.7 g of Cl₂

                     40.5 g of Al -------------------  x

                      x = (40.5 x 212.7) / 53.96

                      x = 8614.35 / 53.96

                      x = 159.64 g of Cl₂

Excess Cl₂ = 212.7 - 159.64

                  = 53.057 g

d) Calculate the moles of Cl

                       35.45 g of Cl ----------------- 1 mol

                       53.057 g of Cl ---------------  x

                        x = (53.057 x 1)/35.45

                       x = 1.49 mol of Cl₂                    

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