Determine the limiting reactant. 4NH3 (g) + 5O2 (g) →4NO (g) + 6H2O (g)

RMM of NH3 = 17

RMM of O2 = 32

By equation,

4 mole of NH3 reacts with 5 mole of O2.

So, (4x17) g of NH3 reacts with (5x32) g of O2

68g of NH3 reacts with 160g of O2

1.5 g of NH3 reacts with? = 160 x 1.5 / 68 = 3.53g of O2

But mass of O2 in the reaction = 2.75 g

So,

(A) Oxygen is the limiting substance.

(B) amount of products are dependent on limiting substance.

RMM of NO = 30

so, 4 mole of NO = 120 g

RMM of H2O = 18

6 moles of H2O = 108

So, 160 g of oxygen produces 120 g of NO

2.75 g of Oxygen = ? = 2.75x 120 / 160 = 2.0625g of NO

also, 160 g of oxygen produces 108 g of H2O

2.75 g of oxygen = ? = 108 x 2.75 / 160 = 1.85625g of H2O

(C) By equation,

160 g of oxygen reacts with 68 g of NH3

2.75 g of oxygen reacts with? = 68 x 2.75 / 160 = 1.16725 of NH3 are consumed.

mass of NH3 excess = 1.5 - 1.16725 = 0.33725 g of NH3