Respuesta :
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
