Question

A chemist combines 11.3 L N2 gas, 34.3 L O2 gas, and 17.5 L Br2 gas...

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

Homework Answers

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

VN2 = 11.3 L

So, Mass of N2 = VN2 * 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

VO2= 34.3 L

So, Mass of O2 = VO2 * 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

VBr2 = 17.5 L

So, Mass of Br2 = VBr2 * 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

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