Question

pleasee i need helpp

f 75.5 g N2 react with 15.4 g H2 according to the reaction below, how many grams of ammonia (NH3) can be produced and how many grams of the excess reactant will be left over? Be sure to show all of your work.

unbalanced equation: N2 + H2 NH3

Answer 1

balanced=N2 + 3 H2 = 2 NH3

Answer 2

If 75.5 g N2 react with 15.4 g H2 according to the reaction below, how many grams of ammonia (NH3) can be produced and how many grams of the excess reactant will be left over? Be sure to show all of your work.

unbalanced equation: N2 + H2 NH3

i) Balance the equation: \(\large \text{N}_2+3\text{H}_2=2\text{NH}_3\)
ii) Stoichiometry: 1 mole of N2 + 3 moles of H2 yield 2 moles of ammonia, let's put that into grams;
28g \((14g/\text{mol} * 2\text{mol})\) of \(\text{N}_2\) and
6.06g\((1.01g/\text{mol} * 2 * 3\text{mol} \text)\) yields
13.03g\((14g/\text{mol}+1.01g/\text{mol}*3)*3 \text{mol})\) of NH3