 Let's just try one more. Again pause and try and work this out for yourself. We're looking at the ammonia reaction and we have the balanced equation here and we're told that we're reacting 7.1 litres of nitrogen gas at STP standard temperature and pressure and we are to calculate the mass of ammonia produced. So let's again sketch out the calculation path. We know the volume of nitrogen that's 7.1 litres and we're going to use the mole of volume 22.4 litres per mole to convert that into moles of nitrogen. The mole ratio from the equation will give us the moles of ammonia produced and we'll then need to work out the molar mass of ammonia so that we can convert the moles to a final mass of ammonia. So let's try that out. First convert the volume of the nitrogen to moles. We know the volume is 7.1 litres and the molar volume is 22.4 litres per mole. So we line those up and divide by the molar volume. Remember it's the same as multiplying by the inverse. The litres cancel out and that gives us 0.3170 moles of nitrogen gas. Second we'll convert those moles of nitrogen into moles of ammonia and we need the mole ratio. We look at the chemical equation and we see that the mole ratio is 1 mole of nitrogen to 2 moles of ammonia. So we need to multiply our moles of nitrogen by 2. That gives us 0.6339 moles of ammonia. Next we need to convert moles of ammonia to a mass so we need the molar mass of ammonia. We check the periodic table and we can calculate that the molar mass of ammonia in H3 is 17.034 grams per mole. We take the moles of ammonia, multiply by the molar mass, the moles cancel out and that gives us 10.798 grams which we round to 11 grams because we started with a value that had two sig figs. Great. Now remember like anything this gets easier with practice. Make sure you put in the practice so that this aspect of chemistry becomes a simple calculation task rather than a source of stress.