# When a sample of NO(g) (11.75 mol) is placed in 440.0 L reaction vessel at 882.0 K and allowed to come to equilibrium the mixture contains 79.86 grams of N2(g). What is the concentration (mol/L) of N2(g)?

Oct 8, 2014

The concentration of N₂ is 6.480 × 10⁻³ mol/L.

At first glance, this looks like an equilibrium calculation, but it is more like a molarity calculation.

1. Calculate the moles of N₂

${\text{Moles of N"_2 = "79.86 g N"_2 × ("1 mol N"_2)/("28.01 g N"_2) = "2.851 mol N}}_{2}$

2. Calculate the concentration of N₂

["N"_2] = "moles"/"litres" = "2.581 mol"/"440.0 L" = 6.480 ×10^-3"mol/L".