Question

80 g of H2 is reacted with 80 g of O2 to form water. find out the mass of water obtained.which substance is the limiting reagent?

Answer 1

80g O2 is the limiting reagent here.
because from theoretical calculations we get
 2H2       +        O2       =       2 H2O  
  4g                   32g                 36g
  80g                 640g               720g
so we have less amout of O2 than require amount.
that is why O2 is limiting here.
from 32g O2 we get 36g of H2O
         80g O2 we get 90g of H2O

Answer 2

Answer:

2H2 + O2 ===} 2H2O

In Mole,

2 mol H2 + 1 mol O2 ===} 2 mol H2O.

In gram,

4g H2 +32g O2 ===} 36g H2O.

Here, 80 g H2 react with 80 g O2 to form water.

So,

Number of moles,

H2 = 80÷2 = 40mol

O2 = 80 ÷ 32 = 2.5mol

Hence, O2 is limiting reagent as number of moles are less than H2.

From reaction,

32 g O2 gives 36 g water.

So, by given data,

80 g O2 gives 80×36÷32 = 90g.

Hense Mass of water obtained is 90g.