Answer :
The moles of nitrogen that would react with excess hydrogen to produce 520 ml of ammonia is 0.011 moles
calculation
step 1: write the balanced equation for reaction
= N2 + 3 H2 → 2NH3
step 2: find the moles of NH3 produced
- convert 520 Ml into liters = 520/1000= 0.52 L
- At STP 1 moles of gas = 24 L
? = 0.52 L
- by cross multiplication
(0.52 L x 1 mole) / 24 L =0.02167 moles of NH3
step 3: use mole ratio to determine the moles of Nitrogen
the mole ratio of N2:NH3 = 1:2
therefore the moles of N2= 0.02167 x1/2=0.011 moles of N2