Given reaction 2NH3(g) + 3Cl2(g) → N2(g) + 6HCl(g), you react 5.0 L of NH3 with 5.0 L of Cl2 measured at the same conditions in a closed container. Calculate the ratio of pressures in the container (Pfinal/Pinitial).

0.75
1.00
1.50
1.33
None of these choices are correct

n

Well, well, well! It looks like we have a fun reaction going on here! Now, to calculate the ratio of pressures, we need to remember a little something called the ideal gas law. The ideal gas law states that the pressure of a gas is directly proportional to its number of moles and its temperature, while inversely proportional to its volume.

Let's start by finding the number of moles of NH3 and Cl2. Since we have 5.0 L of each gas, which are measured at the same conditions, we can assume they have the same number of moles. Since 1 mole of NH3 reacts with 3 moles of Cl2, we will get 5.0/2 = 2.5 moles of NH3 and 2.5 * 3 = 7.5 moles of Cl2 after the reaction.

Now, let's consider the ratio of pressures. According to the balanced equation, the number of moles of NH3 decreases to form the products, while the number of moles of HCl increases. Since the total number of moles increases, the pressure in the container will also increase.

Therefore, the ratio of pressures in the container is greater than 1. None of the given choices is correct. Let's just say there's no clowning around with this question!

To calculate the ratio of pressures in the container (Pfinal/Pinitial), we need to use the concept of stoichiometry and the ideal gas law.

First, let's determine the initial and final number of moles of each gas involved in the reaction:

Given:
Initial volume of NH3 (Vinitial) = 5.0 L
Initial volume of Cl2 (Vinitial) = 5.0 L

Using the ideal gas law, we can calculate the initial number of moles for each gas:

ninitial(NH3) = (PVinitial)/(RT)
Assuming the conditions are constant, we can simplify the equation to:
ninitial(NH3) = (Pinitial)(Vinitial)/(RT)

Similarly, we can calculate the initial number of moles of Cl2:
ninitial(Cl2) = (Pinitial)(Vinitial)/(RT)

Since the stoichiometric coefficient of NH3 is 2, the number of moles of NH3 used in the reaction will be twice the number of moles of N2 formed:
ninitial(NH3) = 2nfinal(N2)

Since the stoichiometric coefficient of Cl2 is 3, the number of moles of Cl2 used in the reaction will be three times the number of moles of HCl formed:
ninitial(Cl2) = 3nfinal(HCl)

We can solve these equations to find the number of moles of N2 and HCl produced in the reaction.

Now, let's consider the final conditions. The number of moles of N2 and HCl in the mixture is the same as the number of moles formed in the reaction. However, the total volume of the mixture will be the sum of the initial volumes of NH3 and Cl2.

Therefore, the final pressure (Pfinal) can be calculated using the ideal gas law:

Pfinal = (nfinal(N2) + nfinal(HCl))RT / (Vinitial(NH3) + Vinitial(Cl2))

Now, we can calculate the ratio of pressures:

Pfinal / Pinitial = [(nfinal(N2) + nfinal(HCl))RT / (Vinitial(NH3) + Vinitial(Cl2))] / [(Pinitial)(Vinitial)/(RT)]

Simplifying the equation, we find:

Pfinal / Pinitial = [(nfinal(N2) + nfinal(HCl))(Vinitial(NH3) + Vinitial(Cl2))] / [(Pinitial)(Vinitial)]

Finally, we substitute the given values into the equation and calculate the ratio of pressures.

identify with resason oxidizing 3cuo (5) *2nh3 (g) 3cu (5)*3 h20 (1)*n2 (g)

2NH3 + 3Cl2 ==> N2 + 6HCl

Determine the limiting reagent. It is Cl2. Repost if you don't know how to do that. So all of the Cl2 will be used up.
How much N2 is formed?
5L x 1/3 = 1.67
How much HCl is formed?
5L x6/3 = 10 L
How much NH3 is used?
5L x 2/3 = 1.67 so how much NH3 is left? 5.0-1.67 = 3.33
So total L = 3.33 + 0 + 1.67 + 10 = 13.34. I have used L in all cases and that substitutes for mols in gas problems.
Total P initial = 5+5 = 10
Total P final = 13.34
You can do the ratio.