if 2.2g of h2 and 1.25g of n2 are present, how many grams of nh3 will be produced

Please try to get into the habit of using the correct symbols as this aids your communication on the topic. (I will assume here that the reaction is going to completion)

Start from a balanced equation

3H2 + N2 -> 2NH3

then calculate the number of moles of each of the starting materials

for H2 2.2 g / 2 g mol^-1

for N2 1.25 g / 28 g mol^-1

now decide from the equation which of the two is in excess, then base your calculation of the amount of product on the starting material that is not in excess.