Ammonia, nh3, is produced industrially from nitrogen and hydrogen. what mass of the reagent in excess will remain when 30.0 g of n2 and 10.0 g of h2 react until the limiting reagent is completely consumed? report your answer to the appropriate number of significant figures.

3.60 grams of Hydrogen

Explanation:

This is a problem of stoichiometry so we first need the balance chemical reaction and then convert the quantities given to mol and determine our limiting reagent to calcuale the amount of the excess reagent if any.

N₂ + 3 H₂   2 NH₃

MW N₂ = 28.0 g/mol  mol N₂ = 30g x 1mol/28.0 g = 1.07 mol

MW H₂ =   2.01 g/mol   molH₂=   10 g x 1 mol /2.01 = 4.98 mol

mol H₂ required to completely react with the N₂ :

3mol H₂/ 1 mol N₂   x 1.07 mol N₂ =  3.21 mol H2

we have plenty of H₂.

mol N₂ required to completely react with the H₂ :

1 mol N₂/ 3 mol H₂ x   4.98 mol H₂ =  1.66 mol N₂

but we have only 1.07 mol N₂, therefore our limiting reagent is nitrogen and hydrogen is in excess. The amount in excess is:

5 mol H₂ -  3.21 mol H₂ reacted =  1.79 mol H₂

1.79 mol H₂ times 2 g/mol will give us the amount  of hydrogen left unconsumed:

1.79 mol x 2.01 g/mol = 3.60g H₂

it's all four options.