3g of H2 react with 29g of O_{2 }to yield H_{2}O .

(1) Which is limiting raegent?

(2) Calculate the maximum amount of H_{2}0 that can be formed?

(3) Calculate the amount of one of the reactants which remains unreacted.

The balanced equation for the above reaction is as follows:

From the above equation it is clear that 2 mole H_{2} react with 1 mole O_{2}

Molar mass of H_{2} = 2g

Molar mass of O_{2}= 32 g

This implies,

4 g H_{2} react with 32 g O_{2}

3 g H_{2} reacts with = (32/4) x 3g of O_{2} gas

= 24 g

As the given amount of O_{2} is more than required therefore O_{2}_{ }is excess reagent and H_{2} is limiting reagent.

B.

2 mole of hydrogen gas reacts to form 2 mole of water molecule therefore,

4 g of H_{2} produces = 36 g of water

So the amount of H_{2}O produced by 3 g H_{2} = (36/4) x 3

= 27 g

Hence, 27 g of water will be produced during the recation

C.

As, 24 g of oxygen has been utilised during the recation and 29 g of oxygen was supplied therefore amount of oxygen gas left is (29-24) = 5g

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