Given 6.0 mol of n2 are mixed with 12.0 mol of h2. equation: n2 + 3h2--> 2nh3

which chemical is in excess? what is the excess in moles?
theoretically, how many moles of nh3 will be produced?
if the percentage yield of nh3 is 80%, how many moles of nh3 are actually produced?

1) Excess reagent

1 mol N2 / 3 mol H2

6.0 mol N2 *3 mol H2 / 1 mol N2 = 18 mol H2

18mol H2 > 12 mol H2 =>  H2 is limiting (you  need 18 mol H2 to use all the 6 mol N2), then N2 is in excees.

12.0 mol H2 * 1mol N2/ 3 mol H2  = 4 mol N2 is the quantity that will react, then the excess is 6 mol N2 - 4 mol N2 = 2 mol N2

2) NH3 produced

12 mol H2 * [2 mol NH3 / 3 mol H2] = 8 mol NH3

Aslso, 4 mol N2 *[2molNH3 / 1 molN2] = 8 mol NH3, the same result.

3) Yield

80% * 8 mol NH3 = 6.4 mol NH3

Olá. porque a reação na temperatura é muito rapida a do copo : )