Answer: The mass of
produced is, 45.8 grams.
Explanation : Given,
Mass of
= 18.0 g
Mass of
= 23.0 g
Molar mass of
= 23 g/mol
Molar mass of
= 71 g/mol
Molar mass of
= 58.5 g/mol
First we have to calculate the moles of
and
.
![\text{Moles of }Na=\frac{\text{Given mass }Na}{\text{Molar mass }Na}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DNa%3D%5Cfrac%7B%5Ctext%7BGiven%20mass%20%7DNa%7D%7B%5Ctext%7BMolar%20mass%20%7DNa%7D)
![\text{Moles of }Na=\frac{18.0g}{23g/mol}=0.783mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DNa%3D%5Cfrac%7B18.0g%7D%7B23g%2Fmol%7D%3D0.783mol)
and,
![\text{Moles of }Cl_2=\frac{\text{Given mass }Cl_2}{\text{Molar mass }Cl_2}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DCl_2%3D%5Cfrac%7B%5Ctext%7BGiven%20mass%20%7DCl_2%7D%7B%5Ctext%7BMolar%20mass%20%7DCl_2%7D)
![\text{Moles of }Cl_2=\frac{23.0g}{35.5g/mol}=0.648mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DCl_2%3D%5Cfrac%7B23.0g%7D%7B35.5g%2Fmol%7D%3D0.648mol)
Now we have to calculate the limiting and excess reagent.
The balanced chemical equation is:
![2Na+Cl_2\rightarrow 2NaCl](https://tex.z-dn.net/?f=2Na%2BCl_2%5Crightarrow%202NaCl)
From the balanced reaction we conclude that
As, 2 mole of
react with 1 mole of ![Cl_2](https://tex.z-dn.net/?f=Cl_2)
So, 0.783 moles of
react with
moles of ![Cl_2](https://tex.z-dn.net/?f=Cl_2)
From this we conclude that,
is an excess reagent because the given moles are greater than the required moles and
is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of ![NaCl](https://tex.z-dn.net/?f=NaCl)
From the reaction, we conclude that
As, 2 mole of
react to give 2 mole of ![NaCl](https://tex.z-dn.net/?f=NaCl)
So, 0.783 mole of
react to give 0.783 mole of ![NaCl](https://tex.z-dn.net/?f=NaCl)
Now we have to calculate the mass of ![NaCl](https://tex.z-dn.net/?f=NaCl)
![\text{ Mass of }NaCl=\text{ Moles of }NaCl\times \text{ Molar mass of }NaCl](https://tex.z-dn.net/?f=%5Ctext%7B%20Mass%20of%20%7DNaCl%3D%5Ctext%7B%20Moles%20of%20%7DNaCl%5Ctimes%20%5Ctext%7B%20Molar%20mass%20of%20%7DNaCl)
![\text{ Mass of }NaCl=(0.783moles)\times (58.5g/mole)=45.8g](https://tex.z-dn.net/?f=%5Ctext%7B%20Mass%20of%20%7DNaCl%3D%280.783moles%29%5Ctimes%20%2858.5g%2Fmole%29%3D45.8g)
Therefore, the mass of
produced is, 45.8 grams.