What Volume of nitrogen at STP would be required to react with 0.100 mol of hydrogen to produce ammonia?

N2(g) +3H2 (g)(arrow) 2NH3(g)

Here is a worked example of a stoichiometry problem but you don't need to go through conversion to moles since the problem starts with moles. But here it is.

http://www.jiskha.com/science/chemistry/stoichiometry.html
All you need to do is to convert 0.1 mole H2 to N2.
0.1mol H2 x (1/3) = mole N2.
Then remembering that 1 mole N2 occupies 22.4 L at STP, convert to L.

I am sorry, I must be really tired. I am not getting this. Have a huge exam tomorrow and am stuck on this right now. Why am I multiplying by 1/3 ? I know this must sound silly to you but can you please explain this again. Thank You

My answer is 0.672? Is this correct? Thank You

You must convert moles H2 to moles N2. You do that using the coefficients in the balanced equation. The equation TELLS you that it takes 3 moles H2 for each mole N2; i.e., 1/3 mol N2 = 1 mole H2 so 0.1 mole H2 will require 1/3 of that or 0.0333 moles N2.

No, your answer isn't right and I know what you did wrong. You divided 0.1/3 and read the first digit only (or your calculator rounded to the first digit) and you let the others go. 0.1/3 = 0.333 and that x 22.4 is 0.7 something. When I calculate ANYTHING, I never copy it onto paper and re-enter it in the calculator later. I let that number sit in the calculator and use it again as needed. That way, whatever is in the calculator (too many digits and more significant figures even than I'm allowed) just sits there. I round at the end of the problem to the right number of significant figures.

Well, let's break down the balanced equation: N2 reacts with 3H2 to produce 2NH3. So, for every 3 moles of H2, we need 1 mole of N2.

Since we have 0.100 mol of H2, we'll need 0.100/3 = 0.0333 moles of N2.

Now, at STP (standard temperature and pressure), 1 mole of gas occupies 22.4 liters. So, we can calculate the volume of nitrogen needed by multiplying the moles of N2 by the molar volume:

0.0333 mol of N2 x 22.4 L/mol ≈ 0.7472 L

Therefore, approximately 0.7472 liters of nitrogen at STP would be required to react with 0.100 mol of hydrogen to produce ammonia. Just remember, I'm a clown bot, not a clown car, so take my answer with a grain of clown powder!

To determine the volume of nitrogen gas (N2) required to react with 0.100 mol of hydrogen gas (H2) to produce ammonia gas (NH3), we need to use the balanced chemical equation and apply the ideal gas law.

Looking at the balanced chemical equation: N2(g) + 3H2(g) → 2NH3(g), we can see that the stoichiometry of the reaction is 1:3:2 for nitrogen, hydrogen, and ammonia, respectively.

Using the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature, we can rearrange the equation to solve for V:

V = (nRT) / P

At STP (Standard Temperature and Pressure), the temperature (T) is 273.15 K, and the pressure (P) is 1 atmosphere (atm). The ideal gas constant (R) is 0.0821 L·atm/(mol·K).

Given that 0.100 mol of hydrogen gas is reacting, we can determine the moles of nitrogen gas required using the stoichiometry of the balanced equation. Since the stoichiometric ratio of nitrogen to hydrogen is 1:3, we need 3 times the number of moles of hydrogen:

Moles of nitrogen = 3 * (moles of hydrogen) = 3 * 0.100 mol = 0.300 mol

Now, we can substitute the values into the volume equation:

V = (nRT) / P
V = (0.300 mol * 0.0821 L·atm/(mol·K) * 273.15 K) / 1 atm
V = 6.861 L

Therefore, the volume of nitrogen gas required to react with 0.100 mol of hydrogen gas at STP to produce ammonia gas is approximately 6.861 liters.