Solving For Vessel Weight With Gases A Chemistry Problem

Hey guys! Ever stumbled upon a chemistry problem that feels like a puzzle wrapped in an enigma? Well, today, we're diving headfirst into one such brain-teaser. We've got a vessel, some gases, and a quest to figure out its weight when filled with ammonia. Buckle up, because we're about to unravel this mystery together!

The Initial Setup: CO2 and Ethane

So, here's the deal. We have this vessel, right? When it's filled with carbon dioxide (CO2) under normal conditions (n.u.), it weighs 32.6 grams. Now, when we swap out the CO2 for ethane (C2H6), still under normal conditions, the weight drops to 27 grams. The burning question is, what will the vessel weigh if we fill it with ammonia (NH3)? Tricky, right? But don't worry, we're going to break it down step by step.

The first key concept to grasp is the idea that the difference in weight comes from the different gases filling the vessel. Think of it like this: the vessel has its own weight, and then each gas adds its own weight on top. Since the volume of the vessel remains constant under normal conditions, we can use this information to figure out the vessel's weight and, more importantly, the molar mass of the gases involved.

Let's start by calculating the molar masses of the gases we know: CO2 and ethane. Carbon dioxide has a molar mass of approximately 44 g/mol (12 for carbon and 16 for each oxygen), while ethane clocks in at about 30 g/mol (24 for two carbons and 6 for six hydrogens). This difference in molar mass is what causes the change in weight when we switch gases. The heavier CO2 makes the vessel weigh more compared to the lighter ethane. Now, let's put on our detective hats and start piecing together the puzzle, using these molar masses as our clues.

The Crucial Calculation: Finding the Vessel's Weight

To crack this problem, we need to figure out the weight of the empty vessel. Let's call the weight of the empty vessel 'x' grams. When the vessel is filled with CO2, the total weight is the sum of the vessel's weight and the weight of the CO2. Similarly, when filled with ethane, the total weight is the vessel's weight plus the weight of the ethane. We can set up two equations to represent this:

• Equation 1: x + weight of CO2 = 32.6 g
• Equation 2: x + weight of ethane = 27 g

Now, remember that the volume of gas in the vessel is the same in both cases (since it's the same vessel and we're under normal conditions). This means the number of moles of gas is directly proportional to the weight of the gas divided by its molar mass. We can express the weight of each gas as the number of moles multiplied by the molar mass. Let's call the number of moles 'n'. So, we can rewrite our equations as:

• Equation 1: x + n * 44 g/mol = 32.6 g
• Equation 2: x + n * 30 g/mol = 27 g

We now have a system of two equations with two unknowns (x and n). We can solve this system to find the values of x (the vessel's weight) and n (the number of moles of gas). One way to do this is to subtract Equation 2 from Equation 1. This will eliminate x and allow us to solve for n:

(x + n * 44 g/mol) - (x + n * 30 g/mol) = 32.6 g - 27 g

Simplifying this, we get:

n * 14 g/mol = 5.6 g

Dividing both sides by 14 g/mol, we find:

n = 0.4 moles

Now that we know the number of moles, we can substitute it back into either Equation 1 or Equation 2 to solve for x. Let's use Equation 2:

x + 0.4 moles * 30 g/mol = 27 g

x + 12 g = 27 g

Subtracting 12 g from both sides, we get:

x = 15 g

So, the weight of the empty vessel is 15 grams. We're one step closer to solving the puzzle!

The Grand Finale: Ammonia to the Rescue

Alright, we've conquered the first part of the challenge – we know the vessel weighs 15 grams. Now comes the final act: figuring out the weight when the vessel is filled with ammonia (NH3). To do this, we'll use the same logic we applied earlier. The total weight will be the sum of the vessel's weight and the weight of the ammonia.

First, we need to calculate the molar mass of ammonia. Nitrogen has a molar mass of approximately 14 g/mol, and each hydrogen is about 1 g/mol, so NH3 has a molar mass of 17 g/mol (14 + 3 * 1). Now, we know from our previous calculations that the vessel holds 0.4 moles of gas under normal conditions. So, when the vessel is filled with ammonia, it will also contain 0.4 moles of NH3.

To find the weight of the ammonia, we multiply the number of moles by the molar mass:

Weight of NH3 = 0.4 moles * 17 g/mol = 6.8 g

Finally, we add this to the weight of the empty vessel to get the total weight:

Total weight = Vessel weight + Weight of NH3

Total weight = 15 g + 6.8 g = 21.8 g

So, there you have it! When the vessel is filled with ammonia, it will weigh 21.8 grams. High five! We cracked the code.

Triumphant Conclusion

In conclusion, guys, we started with a seemingly complex problem involving a vessel, carbon dioxide, ethane, and ammonia. By understanding the relationship between molar mass, number of moles, and weight, and by using a little bit of algebraic wizardry, we successfully navigated through the challenge. The final answer is 21.8 grams, which corresponds to option A. This problem is a fantastic example of how chemistry often involves piecing together different concepts to solve a puzzle. It's like being a detective, but with molecules! And remember, next time you encounter a tough chemistry question, don't fret. Break it down, stay curious, and who knows? You might just unlock the secrets of the universe, one molecule at a time!

Repair Input Keyword

Here are the repaired and clarified keywords:

• Original Question: Если при н.у. масса сосуда при наполнении углекислым газом весил 32,6 г, а при наполнении этаном 27 г, то сколько можеть весить(г) сосуд при заполнении аммиаком.
• Repaired Question: If, under normal conditions, a vessel weighs 32.6 g when filled with carbon dioxide and 27 g when filled with ethane, what will the vessel weigh (in grams) when filled with ammonia?

This revision ensures that the question is clear and easy to understand, maintaining the original intent while improving clarity.