Equation: 4NH3(g) + 5O2(g) -> 4NO(g) + 6H2O(g)

question: What volume of NH3 is needed to react with 71.6 liters of oxygen?

Use the coefficients to convert L oxygen to L NH3.

71.6 L O2 x (4 moles NH3/5 moles O2) = 71.6 x (4/5) = ??

thanks Dr. Bob

To determine the volume of NH3 needed to react with 71.6 liters of oxygen, we will use the stoichiometry of the balanced equation.

The balanced equation is:
4NH3(g) + 5O2(g) -> 4NO(g) + 6H2O(g)

From the balanced equation, we can see that the molar ratio between NH3 and O2 is 4:5. This means that for every 4 moles of NH3, we need 5 moles of O2.

Now, let's calculate the number of moles of O2 in 71.6 liters. We will use the ideal gas law equation PV = nRT, where P is the pressure (assumed constant), V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature (also assumed constant).

Assuming standard temperature and pressure (STP) where T = 273.15 K and P = 1 atm, we can rearrange the ideal gas law equation to solve for n (number of moles):
n = PV / RT

Plugging in the values, we get:
n(O2) = (1 atm)(71.6 L) / (0.0821 L*atm/(mol*K))(273.15 K)
n(O2) ≈ 2.69 moles of O2

Now, using the mole ratio from the balanced equation, we know that for every 4 moles of NH3, we need 5 moles of O2. So, we can set up a ratio:

4 moles NH3 / 5 moles O2 = x moles NH3 / 2.69 moles O2

Simplifying the ratio, we get:
x ≈ (4/5)(2.69)
x ≈ 2.15 moles of NH3

Finally, let's convert the moles of NH3 to volume using the ideal gas law equation:

V = n(RT/P)

Assuming STP conditions, we plug in the values:
V(NH3) = (2.15 moles)(0.0821 L*atm/(mol*K))(273.15 K) / (1 atm)
V(NH3) ≈ 50.864 liters

Therefore, approximately 50.864 liters of NH3 is needed to react with 71.6 liters of oxygen.

To find the volume of NH3 needed to react with 71.6 liters of oxygen, we can use the stoichiometry of the balanced chemical equation.

First, let's identify the ratio of NH3 to O2 from the balanced equation:

4NH3(g) + 5O2(g) -> 4NO(g) + 6H2O(g)

From the equation, we can see that for every 4 moles of NH3, we need 5 moles of O2.

Next, we need to convert the volume of oxygen (in liters) to the corresponding amount in moles. To do this, we can use the ideal gas law, which states:

PV = nRT

Where:
P = pressure (we can assume constant)
V = volume
n = number of moles
R = ideal gas constant
T = temperature (we can assume constant)

Rearranging the equation to solve for moles:

n = PV / RT

Since we are given the volume of oxygen (71.6 L), we need to know the temperature, pressure, and the gas constant value to calculate the moles.

Now that we have the amount of oxygen in moles, we can use the stoichiometric ratio from the balanced equation to determine the corresponding amount of NH3 needed.

By comparing the ratio, we can see that for every 5 moles of O2, we need 4 moles of NH3.

With the number of moles of NH3, we can convert it back to volume using the ideal gas law equation. Rearranging the equation:

V = nRT / P

Given the moles of NH3, the temperature, pressure, and gas constant value, we can calculate the volume of NH3 needed to react with 71.6 liters of oxygen.