Reaction of gases/ gas stoichiometry

[apbuiii]

Homework Statement

The reaction of 50 mL of H2 gas and 50 mL of N2 gas via the equation: 3H2 + N2 ---> 2NH3 will produce how many mL of product? A reaction of 25 mL of Hydrogen gas and 75 mL of Nitrogen gas? Assume that temperature and pressure are constant.

Homework Equations

Avogadro's Law

The Attempt at a Solution

I know that Volume of gases at STP are proportional to the moles of gas present, so the ratios are 3:1:2, and that i mole of ideal gas is 22.4 L. I'm not totally sure as to where to go from here. Is it kind of a limiting reagent because they give me the volumes of each reactant? Please help. Thanks!

 

[Borek]

It is just a simple limiting reagent, with volumes instead of moles - but, as you have already mentioned, in this case moles and volume can be used interchangeably.

 

[apbuiii]

Okay thank you. Well, I asked this question because of this problem: The reaction of 50 mL of chlorine gas with 50 mL of Ethylene gas Cl2 (g) + C2H4 (g) = C2H4Cl2 (g). How many mL of product will be produced? The answer is 50mL but I don't understand why. Since there is 50 mL of chorine and 50 mL of ethylene, shouldn't there be 100 mL of product? But instead, the answer is just 50 mL...

 

[Borek]

Take a look at the reaction equation. How many moles (total) on the left? How many moles on the right?

 

[DeeNos]

I can provide a response to this question by using the principles of gas stoichiometry. First, we need to convert the given volumes of gases (50 mL and 50 mL) into moles using the ideal gas law:

PV = nRT

Where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature. Since the temperature and pressure are constant, we can use the simplified form of the equation, which is n = V/22.4.

For the first reaction, we have:

nH2 = 50 mL/22.4 mL/mol = 2.23 moles
nN2 = 50 mL/22.4 mL/mol = 2.23 moles

Using the balanced chemical equation, we can see that for every 3 moles of H2, we get 2 moles of NH3. Therefore, we have a limiting reagent (H2) and the amount of NH3 produced will be based on the amount of H2 present.

nNH3 = 2.23 moles H2 x (2 moles NH3/3 moles H2) = 1.49 moles NH3

To convert this back to volume, we use the same equation as before:

V = n x 22.4

VNH3 = 1.49 moles x 22.4 mL/mol = 33.4 mL NH3

For the second reaction, we follow the same steps:

nH2 = 25 mL/22.4 mL/mol = 1.12 moles
nN2 = 75 mL/22.4 mL/mol = 3.34 moles

Again, H2 is the limiting reagent and we use the same conversion to find the amount of NH3 produced:

nNH3 = 1.12 moles H2 x (2 moles NH3/3 moles H2) = 0.74 moles NH3

VNH3 = 0.74 moles x 22.4 mL/mol = 16.6 mL NH3

In summary, for the first reaction, 33.4 mL of NH3 will be produced, and for the second reaction, 16.6 mL of NH3 will be produced. It is important to note that these calculations assume ideal gas behavior and may not