#Na_3N# decomposes to form sodium and nitrogen gas at STP. If 13.7 L of nitrogen is produced, how many moles of #Na_3N# was used (22.4 L = 1 mole of any gas) ?

Question

#Na_3N# decomposes to form sodium and nitrogen gas at STP. If 13.7 L of nitrogen is produced,  how many moles of #Na_3N# was used (22.4 L =  1 mole of any gas) ?

Wyatt Adkins · Accepted Answer

#1.22# mol #Na_3N# Given: #Na_3N# decomposes to form sodium (Na) and Nitrogen (#N_2)# at STP (standard temperature and pressure). At STP #22.4 L = 1# mole of gas First you need to write a skeleton equation and then balance it. You need to realize that nitrogen is a diatomic element #N_2#. skeleton equation: #" "Na_3N -> Na + N_2# balanced: #" "2Na_3N -> 6Na + N_2# The balanced equation gives us the mole ratio of sodium nitrate to nitrogen produced: #2/1#. #13.7 cancel(L N_2) xx (1 cancel(mol N_2))/(22.4 cancel(L N_2)) xx (2 mol Na_3N)/(1 cancel(mol N_2)) = 1.22" mol "Na_3N#

Serenity Allen · Answer

undefined To calculate the number of moles of Na₃N used, we first need to find the volume of Na₃N that decomposed to produce 13.7 L of nitrogen gas at STP. Using the given conversion factor of 22.4 L = 1 mole of any gas:

Number of moles of Na₃N = (Volume of nitrogen gas produced) / (22.4 L/mol)

Number of moles of Na₃N = 13.7 L / 22.4 L/mol

Number of moles of Na₃N ≈ 0.6116 moles