Answer:
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
