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Alex73 [517]
3 years ago
8

The equilibrium constant for the decomposition of PCl5 at 250 celcius is 1.05. PCl5(g)-->PCl3(g)+Cl2(g) If the equilibrium pr

essures of PCl5 and PCl3 are .875 atm and .463 atm, respectively, then what is the equilibrium partial pressure of Cl2 at 250 celsius.
Chemistry
1 answer:
mrs_skeptik [129]3 years ago
8 0

Answer : The equilibrium partial pressure of Cl₂ at 25°C is, 1.98 atm

Explanation :

For the given chemical reaction:

PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)

The expression of K_p for above reaction follows:

K_p=\frac{P_{PCl_3}\times P_{Cl_2}}{P_{PCl_5}}

We are given:

P_{PCl_5}=0.875atm

P_{PCl_3}=0.463atm

K_p=1.05

Putting values in above equation, we get:

1.05=\frac{0.463\times P_{Cl_2}}{0.875}\\\\P_{Cl_2}=1.98atm

Thus, the equilibrium partial pressure of Cl₂ at 25°C is, 1.98 atm

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Answer:

2Ag⁺ (aq) + 2OH⁻ (aq) → Ag₂O (s) + H₂O (l)

Explanation:

Step 1: RxN

2AgNO₃ + 2NaOH → Ag₂O + 2NaNO₃ + H₂O

Step 2: Define states of matter

2AgNO₃ (aq) + 2NaOH (aq) → Ag₂O (s) + 2NaNO₃ (aq) + H₂O (l)

Step 3: Total Ionic Equation

2Ag⁺ (aq) + 2NO₃⁻ (aq) + 2Na⁺ (aq) + 2OH⁻ (aq) → Ag₂O (s) + 2Na⁺ (aq) + 2NO₃⁻ (aq) + H₂O (l)

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2Ag⁺ (aq) + 2OH⁻ (aq) → Ag₂O (s) + H₂O (l)

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if the mass of 191 grams NaCl reacted with 74 frams of calcium hydroxide and 80 grams of sodium hydroxide is produced, what mass
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<h3>Answer:</h3>

110.98 g/mol

<h3>Explanation:</h3>

The reaction between NaCl and Ca(OH)₂ is given by the equation;

2NaCl(aq) + Ca(OH)₂(s) → 2NaOH(aq) + CaCl₂(aq)

We are required to determine the mass of CaCl₂ produced,

We will use the following simple steps;

Step 1: Moles of NaCl and Ca(OH)₂ given

Number of moles = Mass ÷ Molar mass

Moles of NaCl

Mass of NaCl = 191 g

Molar mass NaCl = 58.44 g/mol

Number of moles = 191 g ÷ 58.44 g/mol

                             = 3.268 moles

                             = 3.27 Moles

Moles of Ca(OH)₂

Mass of Ca(OH)₂ = 74 g

Molar mass of Ca(OH)₂ = 74.093 g/mol

Number of moles = 74 g ÷ 74.093 g/mol

                             = 0.998 mole

                              = 1.0 mole

However, from the equation  2 moles of NaCl requires 1 mole of Ca(OH)₂

Therefore, from the amount of reactants available NaCl was in excess and Ca(OH)₂ is the limiting reactant .

Step 2: Moles of CaCl₂ produced

From the equation

1 mole of Ca(OH)₂ reacts with NaCl to produce 1 mole of CaCl₂

Therefore; the mole ratio of Ca(OH)₂ to CaCl₂ is 1: 1

Thus;

Moles of CaCl₂ produced is 1.0 moles

Step 3: Mass of CaCl₂ produced

Moles of CaCl₂ = 1.0 mole

Molar mass CaCl₂ = 110.98 g/mol

But; mass = number of moles × Molar mass

Therefore;

Mass of CaCl₂ = 1.0 mole × 110.98 g/mol

                       = 110.98 g CaCl₂

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