A14.01 g sample of n2 reacts with 3.02 g of h2 to form ammonia (nh3). if ammonia is the only product, what mass of ammonia is formed? hint: write the balance equation and use the appropriate mole ratio. it is a limiting reactant problem.

The complete balanced reaction for this is:

N2   +   3H2   -->   2NH3

First we convert the given masses into number of moles.

Molar mass of N2 = 28 g/mol

Molar mass of H2 = 2 g/mol

Therefore,

moles N2 = 14.01 / 28 = 0.5 moles

moles H2 = 3.02 / 2 = 1.51 moles

Then we find which has lower moles/coefficient ratio:

N2 = 0.5 / 1 = 0.5

H2 = 1.51 / 3 = 0.503

Since N2 has lower moles/coefficient ratio, therefore it is the limiting reactant.

So total moles of ammonia formed is:

moles NH3 =  0.5 moles N2 * (2 moles NH3 / 1 mole N2) = 1 mole NH3

The molar mass of NH3 is 17.031 g/mol, hence:

mass NH3 = 1 mole * 17.031 g/mol

mass NH3 = 17.031 grams

oxygen, selenium, tellurium, and polonium.

they're all from the same group as sulfur (group 16).

it is a solid

explanation: