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=(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* 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* 2=9` moles of
`H_2O`

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