Question

A chemist combines 11.3 L N₂ gas, 34.3 L O₂ gas, and 17.5 L Br₂ gas in a reaction vessel at STP. The gases react according to the following equation.

N₂(g) + O₂(g) + Br₂(g) → 2 NOBr(g)

How many moles of each gas are initially present in the vessel, before any reaction occurs?

N2 ___	mol
O2 ___	mol
Br2 ___	mol

Answer 1

Compound	Molar Mass g/mol	Density g/L at 298 K
N2	28.014	1.25
O2	32	1.429
Br2	159.81	3100

V_N2 = 11.3 L

So, Mass of N₂ = V_N2 * density = 11.3 L * 1.25 g/L =14.125 g

Moles of N₂ = Mass /Molar mass= 14.125/28.014 = 0.504 mol

So, Moles of N2 = 0.504 mol

V_O2= 34.3 L

So, Mass of O₂ = V_O2 * density = 34.3 L * 1.429 g/L =49.015 g

Moles of O₂ = Mass /Molar mass= 49.0155/32 = 1.532 mol

So, Moles of O₂ = 1.532 mol

V_Br2 = 17.5 L

So, Mass of Br₂ = V_Br2 * density = 17.5 L * 3100 g/L =54250 g

Moles of Br₂ = Mass /Molar mass= 54250/159.81 = 339.47 mol

So, Moles of Br₂ = 339.47 mol