Question

Calcium oxide reacts with water to produce calcium hydroxide and heat as shown by the equation.

CaO(s) + H2O(l) → Ca(OH)2(aq) + heat
When 32 g of CaO and 14 g of H2O react, how many grams of calcium hydroxide would you expect to be produced? Explain your answer.


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Answer 1

Answer: The mass of
`Ca(OH)_2` produced is, 42.2 grams.

Explanation : Given,

Mass of
`CaO` = 32 g

Mass of
`H_2O` = 14 g

Molar mass of
`CaO` = 56 g/mol

Molar mass of
`H_2O` = 18 g/mol

First we have to calculate the moles of
`CaO` and
`H_2O`.

`\text{Moles of }CaO=\frac{\text{Given mass }CaO}{\text{Molar mass }CaO}`

`\text{Moles of }CaO=(32g)/(56g/mol)=0.571mol`

and,

`\text{Moles of }H_2O=\frac{\text{Given mass }H_2O}{\text{Molar mass }H_2O}`

`\text{Moles of }H_2O=(14g)/(18g/mol)=0.777mol`

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

`CaO(s)+H_2O(l)\rightarrow Ca(OH)_2(aq)+Heat`

From the balanced reaction we conclude that

As, 1 mole of
`CaO` react with 1 mole of
`H_2O`

So, 0.571 moles of
`CaO` react with 0.571 moles of
`H_2O`

From this we conclude that,
`H_2O` is an excess reagent because the given moles are greater than the required moles and
`CaO` is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of
`Ca(OH)_2`

From the reaction, we conclude that

As, 1 mole of
`CaO` react to give 1 mole of
`Ca(OH)_2`

So, 0.571 mole of
`CaO` react to give 0.571 mole of
`Ca(OH)_2`

Now we have to calculate the mass of
`Ca(OH)_2`

`\text{ Mass of }Ca(OH)_2=\text{ Moles of }Ca(OH)_2* \text{ Molar mass of }Ca(OH)_2`

Molar mass of
`Ca(OH)_2` = 74 g/mole

`\text{ Mass of }Ca(OH)_2=(0.571moles)* (74g/mole)=42.2g`

Therefore, the mass of
`Ca(OH)_2` produced is, 42.2 grams.