Respuesta :
Answer: The mass of [tex]NaCl[/tex] produced is, 45.8 grams.
Explanation : Given,
Mass of [tex]Na[/tex] = 18.0 g
Mass of [tex]Cl_2[/tex] = 23.0 g
Molar mass of [tex]Na[/tex] = 23 g/mol
Molar mass of [tex]Cl_2[/tex] = 71 g/mol
Molar mass of [tex]NaCl[/tex] = 58.5 g/mol
First we have to calculate the moles of [tex]Na[/tex] and [tex]Cl_2[/tex].
[tex]\text{Moles of }Na=\frac{\text{Given mass }Na}{\text{Molar mass }Na}[/tex]
[tex]\text{Moles of }Na=\frac{18.0g}{23g/mol}=0.783mol[/tex]
and,
[tex]\text{Moles of }Cl_2=\frac{\text{Given mass }Cl_2}{\text{Molar mass }Cl_2}[/tex]
[tex]\text{Moles of }Cl_2=\frac{23.0g}{35.5g/mol}=0.648mol[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical equation is:
[tex]2Na+Cl_2\rightarrow 2NaCl[/tex]
From the balanced reaction we conclude that
As, 2 mole of [tex]Na[/tex] react with 1 mole of [tex]Cl_2[/tex]
So, 0.783 moles of [tex]Na[/tex] react with [tex]\frac{0.783}{2}=0.392[/tex] moles of [tex]Cl_2[/tex]
From this we conclude that, [tex]Cl_2[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]Na[/tex] is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of [tex]NaCl[/tex]
From the reaction, we conclude that
As, 2 mole of [tex]Na[/tex] react to give 2 mole of [tex]NaCl[/tex]
So, 0.783 mole of [tex]Na[/tex] react to give 0.783 mole of [tex]NaCl[/tex]
Now we have to calculate the mass of [tex]NaCl[/tex]
[tex]\text{ Mass of }NaCl=\text{ Moles of }NaCl\times \text{ Molar mass of }NaCl[/tex]
[tex]\text{ Mass of }NaCl=(0.783moles)\times (58.5g/mole)=45.8g[/tex]
Therefore, the mass of [tex]NaCl[/tex] produced is, 45.8 grams.
