Assuming all gases are at the same temperature and pressure, how many milliliters of hydrogen gas must react to give 41.0 mL of NH3?

(3 / 2) * 41.0

To find the amount of hydrogen gas required to produce 41.0 mL of NH3, we can use the balanced chemical equation and the concept of stoichiometry.

The balanced chemical equation is:

3H2(g) + N2(g) → 2NH3(g)

From the equation, we can see that the stoichiometric ratio between H2 and NH3 is 3:2. This means that for every 3 molecules of H2, we get 2 molecules of NH3.

Now, let's calculate the amount of H2 in moles using the ideal gas law equation:

PV = nRT

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

Since we are given the volume of H2 and NH3 in milliliters, we need to convert them to liters:

41.0 mL = 41.0 mL x (1 L / 1000 mL) = 0.041 L

Now, let's calculate the number of moles of NH3 produced. According to the stoichiometry, 2 moles of NH3 are produced for every 3 moles of H2 reacted:

Moles of NH3 = (moles of H2 / 3) x 2

To find the moles of H2, we can rearrange the ideal gas law equation to solve for n:

n = PV / RT

Assuming that the temperature and pressure are constant, we can use the values to calculate the number of moles of H2:

n (H2) = (P(H2) x V(H2)) / (R x T)

Since the gases are assumed to be at the same temperature and pressure, we can cancel out these terms.

Now, let's substitute the values into the equation:

n (H2) = (P(H2) x V(H2)) / (R x T)

Given that the pressure and volume are not provided, this calculation cannot be done with the given information.