Respuesta :
Answer : The theoretical yield of water formed from the reaction is 0.54 grams.
Solution : Given,
Mass of HCl = 1.1 g
Mass of NaOH = 2.1 g
Molar mass of HCl = 36.5 g/mole
Molar mass of NaOH = 40 g/mole
Molar mass of [tex]H_2O[/tex] = 18 g/mole
First we have to calculate the moles of HCl and NaOH.
[tex]\text{ Moles of }HCl=\frac{\text{ Mass of }HCl}{\text{ Molar mass of }HCl}=\frac{1.1g}{36.5g/mole}=0.030moles[/tex]
[tex]\text{ Moles of }NaOH=\frac{\text{ Mass of }NaOH}{\text{ Molar mass of }NaOH}=\frac{2.1g}{40g/mole}=0.525moles[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
[tex]HCl+NaOH\rightarrow NaCl+H_2O[/tex]
From the balanced reaction we conclude that
As, 1 mole of HCl react with 1 mole of NaOH
So, 0.030 mole of HCl react with 0.030 mole of NaOH
From this we conclude that, [tex]NaOH[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]HCl[/tex] is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of [tex]H_2O[/tex]
From the reaction, we conclude that
As, 1 mole of [tex]HCl[/tex] react to give 1 mole of [tex]H_2O[/tex]
So, 0.030 moles of [tex]HCl[/tex] react to give 0.030 moles of [tex]H_2O[/tex]
Now we have to calculate the mass of [tex]H_2O[/tex]
[tex]\text{ Mass of }H_2O=\text{ Moles of }H_2O\times \text{ Molar mass of }H_2O[/tex]
[tex]\text{ Mass of }H_2O=(0.030moles)\times (18g/mole)=0.54g[/tex]
Therefore, the theoretical yield of water formed from the reaction is 0.54 grams.
