Question

for the reaction ba(oh)2 + 2 hclo3 --> ba(clo3)2+ 2 h2o calculate the number of moles of h2o formed when 0.1 mole of ba (oh)2 is treated with 0.0250 moles of HClO3

Answer 1

Answer : The number of moles of $H_2O$ formed are, 0.0250 mole

Explanation : Given,

Moles of $Ba(OH)_2$ = 0.1 mole

Moles of $HClO_3$ = 0.0250 mole

First we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$Ba(OH)_2+2HClO_3\rightarrow Ba(ClO_3)_2+2H_2O$

From the balanced reaction we conclude that

As, 1 mole of $Ba(OH)_2$ react with 2 moles of $HClO_3$

So, 0.1 moles of $Ba(OH)_2$ react with $\frac{2}{1}\times 0.1=0.2$ moles of $HClO_3$

That means, in the given balanced reaction, $HClO_3$ is a limiting reagent because it limits the formation of products and $Ba(OH)_2$ is an excess reagent.

Now we have to calculate the moles of $H_2O$.

As, 2 moles of $HClO_3$ react to gives 2 moles of $H_2O$

So, 0.0250 moles of $HClO_3$ react to gives $\frac{2}{2}\times 0.0250=0.0250$ moles of $H_2O$

Therefore, the number of moles of $H_2O$ formed are, 0.0250 mole