Assuming all gases are at the same temperature and pressure, how many milliliters of hydrogen gas must react to give 49.0mL of NH3?

3H2(g)+N2(g)→2NH3(g)

Assuming all gases are at the same temperature and pressure, how many milliliters of hydrogen gas must react to give 48.0 mL of NH3?

3H2(g)+N2(g)→2NH3(g)

24mL

To determine the volume of hydrogen gas required to produce 49.0 mL of NH3 gas, we need to use the balanced chemical equation and the stoichiometry (molar ratio) of the reaction.

Looking at the balanced equation:
3H2(g) + N2(g) → 2NH3(g)

We can see that 3 moles of hydrogen gas (H2) react with 1 mole of nitrogen gas (N2) to produce 2 moles of ammonia gas (NH3).

Step 1: Convert the given volume of NH3 to moles.
Given volume of NH3 = 49.0 mL

To convert this volume to moles, we need to use the ideal gas law:

PV = nRT

Where:
P is the pressure (which is assumed to be constant in this case)
V is the volume of gas in liters (convert mL to L by dividing by 1000)
n is the number of moles of gas
R is the ideal gas constant (0.0821 L·atm/mol·K)
T is the temperature in Kelvin (assumed constant)

Let's assume the temperature and pressure are constant and cancel each other out:

P = constant
T = constant
R = constant

Therefore, we can simplify the equation to:

V = n

Converting the given volume of NH3 to liters:
49.0 mL = 49.0 mL ÷ 1000 = 0.049 L

So, the number of moles of NH3 can be calculated as:

n(NH3) = V(NH3) / 22.4 L (since 1 mole of any gas occupies 22.4 L at standard temperature and pressure)

n(NH3) = 0.049 L / 22.4 L

Step 2: Use the stoichiometry of the balanced equation to find the moles of H2.
Since the molar ratio of H2 to NH3 is 3:2, we can use this ratio to find the moles of H2.

n(H2) = n(NH3) × (3 moles H2 / 2 moles NH3)

Step 3: Convert moles of H2 to volume (in mL).
To do this, we need to use the molar volume of a gas at standard temperature and pressure.

1 mole of any gas occupies 22.4 L at standard temperature and pressure.

Therefore, 1 mole of H2 occupies 22.4 L.

So, the volume in milliliters (mL) can be calculated as:

V(H2) = n(H2) × 22.4 L / 1 mole × 1000 mL / 1 L

Now you can substitute the values into the equation and calculate the volume of hydrogen gas (H2) required to produce 49.0 mL of ammonia gas (NH3).