Ammonia may be oxidised to nitrogen monoxide

in the presence of catalyst according to the equation 4 NH3 + 5 O2 gives 4 NO and 6 H2O. If 27 litres of reactants are consumed , what volume of nitrogen monoxide is produced at the same temperature and pressure.

if you started with 27 liters, you will get 10/9 x 27 liters of products.

of that, NO will be .4*10/9*27 liters
and steam will be .6*10/9*27 liters.

if you started with 27 liters, you will get 10/9 x 27 liters of products.

of that, NO will be .4*10/9*27 liters
and steam will be .6*10/9*27 liters.

To determine the volume of nitrogen monoxide (NO) produced, we need to apply the concept of stoichiometry and use the given balanced chemical equation:

4 NH3 + 5 O2 -> 4 NO + 6 H2O

First, let's determine the limiting reactant. The limiting reactant is the one that is completely consumed and determines the maximum amount of product that can be formed.

To find the limiting reactant, we need to compare the number of moles of each reactant to their stoichiometric coefficients. Let's calculate the number of moles for each reactant using the ideal gas law:

PV = nRT

We know the pressure, volume, temperature, and the value of R (the ideal gas constant). From this equation, we can solve for the number of moles (n) for each reactant.

Assuming the temperature and pressure remain constant, the volume of gases is directly proportional to the number of moles. Therefore, we can compare the number of moles to determine the limiting reactant.

Next, we need to calculate the number of moles of NH3 and O2 for the given volume of 27 liters. To do this, we need to use the ideal gas law in the rearranged form:

n = PV / RT

where P is the pressure, V is the volume, R is the ideal gas constant, and T is the temperature.

Let's calculate the moles of NH3:

n(NH3) = (P * V) / (R * T) = (1 * 27) / (0.0821 * T) = 32.95 / T

Similarly, we calculate the moles of O2:

n(O2) = (P * V) / (R * T) = (1 * 27) / (0.0821 * T) = 32.95 / T

Now, let's compare the moles of reactants to their stoichiometric coefficients. From the balanced equation:

4 NH3 : 4 NO

Since the ratio is 1:1, we can conclude that the moles of NH3 will be equal to the moles of NO. Therefore, the moles of NO produced will be (32.95 / T) moles.

Lastly, let's convert the moles of NO to volume using the ideal gas law. Rearranging the ideal gas law equation:

V = nRT / P

We can substitute the value of n (moles of NO) into the equation:

V(NO) = (32.95 / T) * 0.0821 * T / 1

The T cancels out, leaving us with:

V(NO) = 32.95 * 0.0821 / 1

V(NO) = 2.699 liters

Therefore, at the same temperature and pressure, the volume of nitrogen monoxide produced is approximately 2.699 liters.