Question

400 g of N2 and 150 g of H2 are mixed together to form NH3. Identify the limiting reagent and calculate the amount of NH3 produced.

Answer 1

Answer:

limiteing reagents is hydrogen

Explanation:

mole N-400/28=14.28

mole H-150/2=75

14.28 mole of nitrogen react with 75 mole of H2

H2 will extra so H2will be limitin reagent

NH3will formed 14.mole

Answer 2

Answer:

Explanation:

No of moles of n2=400/28=14.2

No of moles of h2=150/2=75

N2 is the limiting reagent

28 g of n2 gives 34 g of nh3

400g gives (34/28)× 400=485 g