Question

How many moles of barium hydroxide can be produced from the reaction of 4.234 moles of barium with 8.902 moles of water according to the following reaction? Ba + 2 H2O Ba(OH)2 + H2

Answer 1

Answer: Thus 4.234 moles of
`Ba(OH)_2` will be produced from the given masses of both reactants.

Step-by-step explanation:

The balanced chemical reaction is :

`Ba+2H_2O\rightarrow Ba(OH)_2+H_2`

According to stoichiometry :

1 mole of
`Ba` require = 2 moles of
`H_2O`

Thus 4.234 moles of
`Ba` will require=
`(2)/(1)* 4.234=8.468moles` of
`H_2O`

Thus
`Ba` is the limiting reagent as it limits the formation of product and
`H_2O` is the excess reagent as it is present in more amount than required.

As 1 mole of
`Ba` give = 1 mole of
`Ba(OH)_2`

Thus 4.234 moles of
`Ba` will give =
`(1)/(1)* 4.234=4.234moles` of
`Ba(OH)_2`

Thus 4.234 moles of
`Ba(OH)_2` will be produced from the given masses of both reactants.