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amm1812
2 years ago
7

In the chemical reaction:Br2(g) + Cl2(g) ⇌ 2 BrCl(g) KP = 0.150If there is initially 0.500 atm of BrCl and nothing else in a con

tainer, what would be the equilibrium concentration of BrCl?
Chemistry
1 answer:
klemol [59]2 years ago
7 0

Answer:

The partial pressure of BrCl at equilibrium is 0.08 atm.

Explanation:

The equilibrium constant of the reaction is given by =K_p=0.150

Br_2(g) + Cl_2(g)\rightleftharpoons 2 BrCl(g)

initial

0       0         0.500 atm

At equilbrium

p      p       (0.500-2p)

The equilibrium constant's expression of the reaction is given by ;

 K_p=\frac{[BrCl]^2}{[Br_2][Cl_2]}

0.150=\frac{(0.500-2p)^2}{p\times p}

Solving for p:

p = 0.21 atm

The partial pressure of BrCl at equilibrium is:

(0.500-2p) = (0.500 - 2 × 0.21 )atm = 0.08 atm[/tex]

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G. H2S contains two hydrogen atoms
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2 years ago
Consider separate solutions of NaOH and KClmade by dissolving 100.0 g of each solute in 250.0 mL of solution. Calculate the conc
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Answer:

Explanation:

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100g NaOH / 40 g = 25 mol

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2 years ago
Hydrogen cyanide, HCN, is prepared from ammonia, air, and natural gas (CH₄), by the following process:
kykrilka [37]

Answer:

6.75 g of HCN can be produced by the reaction

Explanation:

Complete reaction is:

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Let's determine the moles of each reactant:

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12 g . 1 mol / 32g = 0.375 moles of oxygen

10.5 g . 1mol/ 16 g =  0.656 moles of methane

Now is all about rules of three:

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(0.676 . 2) / 2 = 0.676 moles of methane

Both can be the limiting reactant.

3 moles of O₂ react with 2 moles of NH₃ and 2 moles of methane

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The same amount for methane, 0.375 moles

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0.656 moles of methane would react with 0.656 moles of NH₃

(0.656 . 3 ) /2 = 0.437 moles of O₂   I do not have enough O₂

Oxygen is the limiting reactant → We can work with the reaction now.

Ratio is 3:2. 3 moles of oxygen produce 2 moles of cyanide

0.375 moles of O₂ may produce (0.375 .2 ) / 3 = 0.250 moles

If we convert the moles to mass → 0.250 mol . 27 g / 1mol = 6.75 g

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