How much energy is released when 67.04g of phosphorous is reacted with 10.20g of chlorine? ___ P + ___ Cl2 ___ PCl3 ΔH = -574 kJ

Answers

Answer 1

Answer:

26.78 kJ

Explanation:

To solve the problem, we have to first write the stoichiometric coefficients in the chemical equation:

2P + 3Cl₂ → 2PCl₃

With these coefficients, we have the same number of atoms of each chemical element on both sides: 2 atoms of P, 6 atoms of Cl.

According to the equation, 2 moles of phosphorous (P) react with 3 moles of chlorine (Cl₂), and 574 kJ of energy are released. We have to figure out which is the limiting reactant. For this, we convert the mass into moles by using the molar mass(MM):

MM(P) = 30.9 g/mol

67.04 g P/(30.9 g/mol) = 2.17 mol P

MM(Cl₂) = 35.4 g/mol x 2 = 70.8 g/mol

10.20 g Cl₂/(70.8 g/mol) = 0.14 mol Cl₂

Now, we multiply the actual moles of P (the amount we have for the reaction) by the stoichiometric ratio given by the chemical equation (3 mol Cl₂/2 mol P):

2.17 mol P x (3 mol Cl₂/2 mol P) = 3.25 mol Cl₂

To completely react 67.04 g P, we need 3.25 mol of Cl₂, and we have only 0.14 moles of Cl₂, so the limiting reactant is Cl₂.

Now, we use the limiting reactant to calculate the energy released from the reaction. The energy released per mole of Cl₂ is:

ΔH/(3 mol Cl₂) = -574 kJ/3 mol Cl₂= 191.3 kJ/mol Cl₂

Finally, we multiply the energy released per mole of Cl₂ by the number of moles of Cl₂ we have:

0.14 mol Cl₂ x 191.3 kJ/mol Cl₂ = 26.78 kJ


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How is a scientific law different from other laws in society?
A. A scientific law serves to protect people, property, and nature from
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D. A scientific law is what scientists have decided should be forced
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Answer:

B. A scientific law describes what always happens under specific

conditions in nature.

Explanation:

Scientific law is something that has been proven over and over and therefore is always correct under conditions in nature.

Answer:

B. A scientific law describes what always happens under specific

conditions in nature.

Explanation:

Scientific law is based on observation and experiments to support a claim. It usually describes nature.  

Dalton's contribution was different from thatof the ancient Greek who postulated the existence of atoms. Point out the differences between the two ideas​

Answers

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Answers

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Answer:

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Answer:

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Answers

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Answer:

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Note: The question is incomplete. The complete question is given blow.

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Answers

Answer:

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Answers

Answer:

[tex]H_5N_5O_6[/tex]

Explanation:

Hello there!

In this case, for the determination of empirical formulas, it is firstly necessary to assume the given percentages of the constituent atoms as the masses so we can compute their moles in the formula:

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However, we need to turn all these numbers, whole numbers, so we multiply by 5 to get:

[tex]H_5N_5O_6[/tex]

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Answers

Answer:

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Explanation:

All explanations or thoughts led to that answer

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Answers

Answer:

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attached below is the detailed solution

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Answers

Answer:

b) 2H+(aq) + 2C1-(aq) + Zn(s) → H2(g) + Zn2+(aq) + 2Cl-(aq)

Explanation:

The equation is given as;

2HCl(aq) + Zn(s) + H2(g) + ZnCl2(aq)

In writing an ionic equation, only the aqueous compounds dissociates into ions. This means HCl and ZnCl2 would dissociate to form ions.

This is given as;

2H+  +  2Cl-  + Zn(s) --> H2(g) + Zn2+  + 2Cl-

The correct option is;

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Answers

Answer:

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Explanation:

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Answers

Answer:

Explanation:

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Answers

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WHITE BLOOD SELL?

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Answers

Answer:

Zinc mercury transition metal

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Explanation:

1- What is the molar mass of the following compounds?
a. Ca(OH) 2 _______________________________=
b. H 2 O 2 _______________________________=
c. KCl _______________________________=
d. NaH 2 PO_______________________________=

Answers

Answer:

a. Ca(OH)₂ = 74.08 g/mol

b. H₂O₂ = 34 g/mol

c. KCl = 74.55 g/mol

d. NaH₂PO₄  = 119.97 g/mol

Explanation:

In order to determine the molar mass of each compound we need to sum, the molar mass of each element.

a. Ca(OH)

40.08 g/mol (Ca) + 2 . 16 g/mol (O) + 2 . 1 g/mol (H) = 74.08 g/mol

b. HO2 . 1 g/mol (H) + 2 . 16 g/mol (O) = 34 g/mol

c. KCl → 39.1 g/mol (K) + 35.45 g/mol (Cl) = 74.55 g/mol

d. NaHPO

23 g/mol (Na) + 2 . 1 g/mol (H) + 30.97 g/mol (P) + 4 . 16 g/mol (O) = 119.97 g/mol

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