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4 years ago

12.5 g of copper are reacted with an excess of chlorine gas, and 25.4 g of copper(II) chloride are

Chemistry

1 answer:

Alika [10]4 years ago

8 0

Answer:

Percent yield = 94.5%

Theoretical yield = 26.89 g

Explanation:

Given data:

Mass of copper = 12.5 g

Mass of copper chloride produced = 25.4 g

Theoretical yield = ?

Percent yield = ?

Solution:

Cu + Cl₂ → CuCl₂

Number of moles of Copper:

Number of moles = mass/ molar mass

Number of moles = 12.5 g/ 63.55 g/mol

Number of moles = 0.2 mol

Now we will compare the moles of copper with copper chloride.

Cu : CuCl₂

1 : 1

0.2 : 0.2

Theoretical yield:

Mass of copper chloride:

Mass = Number of moles × molar mass

Mass = 0.2 mol × 134.45 g/mol

Mass = 26.89 g

Percent yield:

Percent yield = Actual yield / theoretical yield × 100

Percent yield = 25.4 g/26.89 g × 100

Percent yield = 94.5%

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At standard pressure, the total amount of heat required to completely vaporize a 100. gram sample of water at its boiling point

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3 years ago

Consider the following chemical equilibrium: 4NH3+3O2=2N2+6H2O Now write an equation below that shows how to calculate from for

erma4kov [3.2K]

This is an incomplete question, here is a complete question.

Consider the following chemical equilibrium:

$4NH_3(g)+3O_2(g)\rightleftharpoons 2N_2(g)+6H_2O(g)$

Now write an equation below that shows how to calculate Kp from Kc for this reaction at an absolute temperature T. You can assume T is comfortably above room temperature. If you include any common physical constants in your equation be sure you use their standard symbols, found in the ALEKS Calculator.

Answer :

The expression of $K_c$ will be,

$K_c=\frac{[N_2]^2[H_2O]^6}{[NH_3]^4[O_2]^3}$

The expression of $K_p$ will be,

$K_p=K_c(RT)$

Explanation :

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.

As we know that the concentrations of pure solids and liquids are constant that is they do not change. Thus, they are not included in the equilibrium expression.

The given equilibrium reaction is,

$4NH_3(g)+3O_2(g)\rightleftharpoons 2N_2(g)+6H_2O(g)$

The expression of $K_c$ will be,

$K_c=\frac{[N_2]^2[H_2O]^6}{[NH_3]^4[O_2]^3}$

The relation between $K_p$ and $K_c$ is:

$K_p=K_c(RT)^{\Delta n}$