Question

1.5g of H2 Reacts with 14.5g of O2 to form water
(1) which is the limiting reactant ?
(2) calculate the amount of water that is formed in the reaction ?
(3) calculate the amount of the reactant which remains unreacted ?
anyobe plz answer

Answer 1

( 1). Relative ratio = Moles / Coefficient

Relative ratio of H2 = 1.5 / 1
Relative ratio of O2 = 14.5 / 1

Hence, Relative ratio of H2 is less than O2, So H2 will be the LIMITING REAGENT.

( 2). 2 H2 ( 4 g) + O2 ( 32 g) => 2 H2 O ( 36 g)
1.5 g H2 14.5 g O2 ?? ( H2O)

2 moles of H2 reacts with 1 mole O 2 to give 2 moles of H2O.
4g of H2 reacts with 32 g O2 to give 36 g H2O
1.5 g of H2 will react with 14.5 g of O2 to give 36 / 4 * 1.5 = 13.5 g of H2O

(3). O2 Remains unreacted,
1.5 g of H2 reacts with 32 g of O2

1.5 *32 / 4 = 12. 0 g of O2

14.5 g of O2 was available for reaction and hence, 12.0 g of is consumed in reaction,

So, 14.5 g - 12. 0 g = 2.5 g of O2

Since, 2.5 g of O2 is left unreacted.