Question

505 grams of KOH are required to completely react with 4.50 mole of sulfuric acid. How many moles of products are produced?

Answer 1

Answer : The moles of products $K_2SO_4$ and $H_2O$ are, 4.50 and 9 moles.

Explanation : Given,

Mass of water = 5.2 g

Molar mass of water = 18 g/mole

Molar mass of $O_2$ = 32 g/mole

The balanced chemical reaction will be,

$2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O$

First we have to calculate the moles of KOH.

$\text{Moles of }KOH=\frac{\text{Mass of }KOH}{\text{Molar mass of }KOH}$

$\text{Moles of }KOH=\frac{505g}{56g/mole}=9.018mole$

Now we have to calculate the limiting and excess reactant.

From the balanced reaction we conclude that

As, 1 mole of $H_2SO_4$ react with 2 mole of $KOH$

So, 4.50 moles of $H_2SO_4$ react with $4.50\times 2=9$ moles of $KOH$

From this we conclude that, $KOH$ is an excess reagent because the given moles are greater than the required moles and $H_2SO_4$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of products $K_2SO_4$ and $H_2O$.

From the balanced chemical reaction, we conclude that

As, 1 moles of $H_2SO_4$ react to give 1 moles of $K_2SO_4$

So, 4.50 moles of $H_2SO_4$ react to give 4.50 moles of $K_2SO_4$

and,

As, 1 moles of $H_2SO_4$ react to give 2 moles of $H_2O$

So, 4.50 moles of $H_2SO_4$ react to give $4.50\times 2=9$ moles of $H_2O$

Therefore, the moles of products $K_2SO_4$ and $H_2O$ are, 4.50 and 9 moles.