Respuesta :
Answer: The mass of [tex]Ca(OH)_2[/tex] produced is, 42.2 grams.
Explanation : Given,
Mass of [tex]CaO[/tex] = 32 g
Mass of [tex]H_2O[/tex] = 14 g
Molar mass of [tex]CaO[/tex] = 56 g/mol
Molar mass of [tex]H_2O[/tex] = 18 g/mol
First we have to calculate the moles of [tex]CaO[/tex] and [tex]H_2O[/tex].
[tex]\text{Moles of }CaO=\frac{\text{Given mass }CaO}{\text{Molar mass }CaO}[/tex]
[tex]\text{Moles of }CaO=\frac{32g}{56g/mol}=0.571mol[/tex]
and,
[tex]\text{Moles of }H_2O=\frac{\text{Given mass }H_2O}{\text{Molar mass }H_2O}[/tex]
[tex]\text{Moles of }H_2O=\frac{14g}{18g/mol}=0.777mol[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical equation is:
[tex]CaO(s)+H_2O(l)\rightarrow Ca(OH)_2(aq)+Heat[/tex]
From the balanced reaction we conclude that
As, 1 mole of [tex]CaO[/tex] react with 1 mole of [tex]H_2O[/tex]
So, 0.571 moles of [tex]CaO[/tex] react with 0.571 moles of [tex]H_2O[/tex]
From this we conclude that, [tex]H_2O[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]CaO[/tex] is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of [tex]Ca(OH)_2[/tex]
From the reaction, we conclude that
As, 1 mole of [tex]CaO[/tex] react to give 1 mole of [tex]Ca(OH)_2[/tex]
So, 0.571 mole of [tex]CaO[/tex] react to give 0.571 mole of [tex]Ca(OH)_2[/tex]
Now we have to calculate the mass of [tex]Ca(OH)_2[/tex]
[tex]\text{ Mass of }Ca(OH)_2=\text{ Moles of }Ca(OH)_2\times \text{ Molar mass of }Ca(OH)_2[/tex]
Molar mass of [tex]Ca(OH)_2[/tex] = 74 g/mole
[tex]\text{ Mass of }Ca(OH)_2=(0.571moles)\times (74g/mole)=42.2g[/tex]
Therefore, the mass of [tex]Ca(OH)_2[/tex] produced is, 42.2 grams.
