How many grams of h2 are needed to produce 10.78 g of nh3?

Ammonia is formed by a reaction between hydrogen and nitrogen as shown by the equation below.

N2 (g) + 3H2 (g) = 2NH3 (g)

1 mole of ammonia contains 17 g

Therefore 10.78 g of ammonia are equivalent to 10.78/17 = 0.6341 moles

The mole ratio of hydrogen to ammonia is 3 : 2

Therefore, moles of hydrogen used will be 0.6341 * 3/2 = 0.9512 moles

1 mole of hydrogen is equivalent to 2 g

Thus, the mas of hydrogen will be 0.9512 moles * 2 = 1.9023 g