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Ch.10 - Gases

Chapter 10, Problem 122b

Ammonia and hydrogen chloride react to form solid ammonium chloride: NH3(g) + HCl(g) NH4Cl(s)

Two 2.00-L flasks at 25 °C are connected by a valve, as shown in the drawing. One flask contains 5.00 g of NH3(g), and the other contains 5.00 g of HCl(g). When the valve is opened, the gases react until one is completely consumed. (b) What will be the final pressure of the system after the reaction is complete? (Neglect the volume of the ammonium chloride formed.)

Two flasks labeled NH3 and HCl connected by a valve, illustrating gas stoichiometry.

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Welcome back everyone to another video ammonia and hydro chloride react to form solid ammonium chloride. According to the reaction, NH three gas plus HCL gas form NH four cl solid 2 2 L flasks at 25 °C are connected by a wave. As shown in the drawing. One flask contains 5 g of ammonia and the other contains 3 g of HCL gas. When the valve is opened, the gasses react until one is completely consumed. What will be the final pressure of the system after the reaction is complete neglect the volume of the ammonium chloride formed. And we are given for answer choices. A 0.67 B 1.30 C 1.98 and D 3.65. All of them are given in atmospheres. So let's solve this problem. Well, we're given our balance reaction equation. There's nothing that we need to do. We can just rewrite it. There's no necessity to include the states at the moment because they're already given to us. We just want to clearly see what's happening. So we have a 1 to 1 reaction, one mole of ammonia reacts with one mole of HCL. What we want to do is just identify the number of moles of each gas initially before the reaction begins. So, starting with NH three, we have 5 g and we need to divide that by the molar mass of ammonia to get the number of moles. We're taking 17 g per mole. And we can say that the number of moles of ammonia would be 0.2941. Now, for the number of moles of hydrogen chloride, we're taking 3 g given to us initially. And we are dividing by the molar mass. In this case, that would be 36 0.5 g per more. If we perform the math, we get 0.0822 moles. Now because the reaction is 1 to 1, we're looking for the limiting reactant and we can say that the limiting reactant would be ac L why? Well, essentially because it's a lower amount and we have a 1 to 1 ratio. Now, if HCL is the limiting reactant, what we can do is is, is find the excess amount of ammonia is the 1 to 1 reaction. Meaning if we subtract 0.29 for one moles and 0.0822 moles, we know that 0.0822 moles will also be equal to the amount of ammonia reacted. So we are finding an excess amount. In this case, we end up with 0.2119 moles. So we know that when the two gasses react, we completely eliminate hydro chloride. In this case, it says H PR should state HCL, right. So we're completely eliminating, eliminating hydro chloride because it completely reacts and we still have some excess amount of ammonia. Now, the problem also suggests that initially, we have 2 L for each part. Meaning when we open the valve, the total volume would be equal to 4 L. That's really important for us because we are already thinking about the Ideal Gas Law PV equals N RT. And we're looking for pressure, P equals N RT divided by V. We have the number of moles of gas remaining. We know the temperature, we know the final volume. So we are ready to solve the problem and state that pressure equals 0.2119 moles multiplied by R the universal gas constant, 0.08206 L multiplied by atmospheres divided by mole multiplied by Calvin. Now, our temperature is 25 °C. We want to convert that into Kelvin. So we are adding 273.15 Kelvin. Now essentially what we want to do is just divide this by volume, which in this case is 4 L. Now the answer we get is 1.30 atmospheres. Looking at the answer choices, we can say that B is the correct answer. Now, that will be it for today. And Thank you for watching.
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