Ammonia gas (NH3) combines with oxygen gas (O2) to form diatomic nitrogen gas and water vapor. If 4.0 grams of ammonia react, how many liters of nitrogen gas are produced? The gas is collected at 32.00°C and a pressure of 2.6 atmospheres.

1.13 L

Explanation:

First, we have to write the chemical equation for the reaction between ammonia gas (NH₃) and oxygen gas (O₂) to give nitrogen gas (N₂) and water (H₂O), as follows:

NH₃(g) + O₂(g) → N₂(g) + H₂O(g)

Then, we have to balance the equation (we write first the coefficient 2 for NH₃ to balance N atoms, then a coefficient of 3 for H₂O to balance H atoms, and finally 1/3 to balance the O atoms):

2 NH₃(g) + 3/2 O₂(g) → N₂(g) + 3 H₂O(g)

In the balanced equation, we can see that 1 mol of N₂ is produced from 2 moles of NH₃. We convert the moles of NH₃ to grams by using its molecular weight (MW):

MW(NH₃) = 14 g/mol N x 1 + (1 g/mol H x 3) = 17 g/mol

grams of NH₃ = 17 g/mol x 2 = 34 g

Thus, we have the stoichiometric ratio:

1 mol of N₂/2 mol NH₃ = 1 mol of N₂/34 g NH₃

To calculate how many moles of N₂ are produced from 4.0 of NH₃, we multiply the mass by the conversion factor:

4.0 g NH₃ x 1 mol of N₂/34 g NH₃ = 0.1176 moles N₂

Finally, we calculate the liters of N₂ gas by using the ideal gas equation:

PV = nRT ⇒ V = nRT/P

We introduce the data in the equation:

T = 32.00°C + 273 = 305 K

P = 2.6 atm

R = 0.082 L.atm/K.mol (is the gas constant)

n= 0.1176 moles

⇒ V = nRT/P = (0.1176 mol x 0.082 L.atm/K.mol x 305 K)/(2.6 atm)

                     = 1.13 L