2.00 L of NO(g) at 1.00 atm and 25°C is reacted with 1.00 L of H2(g) at 2.00 atm and 25°C according to the following unbalanced equation.

NO(g) + H2(g) ® N2(g) + H2O(g)
Which, if either, is the limiting reactant?

When all are gaseous, one may use L as if L = moles, a good short cut to remember. Second, we must make sure both are at the same pressure; that can be accomplished by changing H2 to 1.00 atm. 1L H2 @ 2.00 atm = 2 L H2 @ 1.00 atm.

2NO(g) + 2H2(g) ® N2(g) + 2H2O(g)

Use the coefficients in the balanced equation to convert L NO and L H2 to L of either of the products.
2L NO x (1 mole N2/2 mols NO) = 2(1/2) = 1 mole N2 formed.

2L H2 X (1 mole N2/2 moles H2) = 2(1/2) = 1 mol N2 produced.
Therefore, neither is the limiting reagent. Or we could say both are limiting reagents.

Oh, it's time for some chemical clowning!

Let's see, we're talking about limiting reactants here. Think of it like having a monopoly on the reaction party. You want to know who's going to run out first and ruin the fun!

To figure it out, we'll need to convert the volumes to moles. So, let's grab our calculators, nerdy math hats on, and get calculating.

Using the ideal gas law (because we fancy like that), we can convert the volume to moles:

For NO(g):
P = 1.00 atm
V = 2.00 L
n = PV/RT

Look at that, we calculated n. Give me a drumroll, please! *Ba dum tss* You have moles of NO(g)!

Now, onto H2(g):
P = 2.00 atm
V = 1.00 L
n = PV/RT

We did it again, folks! We have moles of H2(g)!

Now comes the part where the limiting reactant reveals itself. *Cue suspenseful music*

We need to compare the mole ratios from the unbalanced equation. I'll save you the anxiety and do the math for you:

1 mole of NO(g) reacts with 1 mole of H2(g)

And the winner is... the reactant with the smallest mole ratio!

So, to sum it up, whichever reactant has less moles is the limiting reactant. The one that runs out first and kills the party.

I hope this hasn't been too much of a Gas-gastronomical circus act! If you have any more questions, remember, I'm your trusty Clown Bot!

To determine the limiting reactant, we need to calculate the number of moles of each reactant and compare their stoichiometric ratios.

First, let's calculate the number of moles of NO(g). We can use the ideal gas law equation to calculate the number of moles:

PV = nRT

Where:
P = pressure (in atm)
V = volume (in L)
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)

Given:
P(NO) = 1.00 atm
V(NO) = 2.00 L
T(NO) = 25°C = 25 + 273.15 = 298.15 K

Substituting the values into the ideal gas law equation for NO:

(1.00 atm) * (2.00 L) = n(NO) * (0.0821 L·atm/mol·K) * (298.15 K)

Simplifying the equation:

2.00 = n(NO) * 24.495

Dividing both sides by 24.495:

n(NO) = 2.00 / 24.495
n(NO) ≈ 0.0818 mol

Next, let's calculate the number of moles of H2(g) using the same procedure:

Given:
P(H2) = 2.00 atm
V(H2) = 1.00 L
T(H2) = 25°C = 25 + 273.15 = 298.15 K

Using the ideal gas law equation for H2:

(2.00 atm) * (1.00 L) = n(H2) * (0.0821 L·atm/mol·K) * (298.15 K)

Simplifying the equation:

2.00 = n(H2) * 24.495

Dividing both sides by 24.495:

n(H2) = 2.00 / 24.495
n(H2) ≈ 0.0818 mol

Now, let's compare the moles of NO and H2. We see that both reactants have approximately the same number of moles: 0.0818 mol.

According to the balanced equation, the stoichiometric ratio between NO and H2 is 1:1. This means that for every 1 mol of NO reacted, we also need 1 mol of H2.

Since both reactants have the same number of moles, they are in a 1:1 ratio. This means that neither reactant is in excess, and both are completely consumed during the reaction.

Therefore, neither NO nor H2 is the limiting reactant in this case.

To determine the limiting reactant in a chemical reaction, you need to compare the stoichiometric ratios of the reactants and their available quantities. In this case, the given quantities of NO(g) and H2(g) at given temperature and pressure are:

NO(g): 2.00 L, 1.00 atm
H2(g): 1.00 L, 2.00 atm

To make a valid comparison, we need to convert the given volumes to the number of moles of each gas using the Ideal Gas Law: PV = nRT. The equation for the reaction shows that 1 mole of NO reacts with 1 mole of H2, producing 1 mole of N2 and 1 mole of H2O.

First, let's calculate the number of moles for each gas:

NO(g):
n_NO = (P_NO * V_NO) / (R * T)
= (1.00 atm * 2.00 L) / (0.0821 L•atm/mol•K * (25°C + 273.15))
≈ 0.0943 moles of NO

H2(g):
n_H2 = (P_H2 * V_H2) / (R * T)
= (2.00 atm * 1.00 L) / (0.0821 L•atm/mol•K * (25°C + 273.15))
≈ 0.0803 moles of H2

Now that we have determined the number of moles for each gas, we can compare the stoichiometric ratio based on the balanced equation. Since the stoichiometric ratio of NO to H2 is 1:1, the limiting reactant will be the one with the smaller number of moles. Therefore, in this case, the limiting reactant is H2(g) with approximately 0.0803 moles.

Please note that the above calculation assumes an ideal gas behavior and uses the ideal gas constant (R = 0.0821 L•atm/mol•K). Additionally, ensure that the temperature is in Kelvin (K) for the calculations.