Question

If 0.3 moles of N2 were reacted with 0.6 moles of H2, what mass of NH3 could form? N2 +3H2 -------> 2 NH3

a) 0.6 g b) 1.2 g c) 6.8 g d) 10.2 g

I know how to do the math portion, I'm just confused on how to determine the limiting and excess reagent. a) 0.6 g b) 1.2 g c) 6.8 g d) 10.2 g

Answer 1

Balanced equation is

N2 +3H2 -------> 2 NH3

Fromm the balanced equation we can say that

1 mole of N2 requires 3 mole of H2 so

0.3 mole of N2 will require

= 0.3 mole of N2 *( 3 mole of H2 / 1 mole of N2)

= 0.9 mole of H2 will be required

But we have 0.6 mole of H2 which is in short so H2 is limiting reactant

From the balanced equation we can say that

3 mole of H2 produces 2 mole of NH3 so

0.6 mole of H2 will produce = 0.4 mole of NH3 is formed

mass of 1mole of NH3 = 17.0 g

Therefore, the mass of 0.4 mole of NH3 = 6.8 g

Therefore, option C is correct