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.

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=$\frac{2}{1}\times 4.234=8.468moles$  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 =$\frac{1}{1}\times 4.234=4.234moles$  of $Ba(OH)_2$

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