Question

2Na+Cl 2 ​ →2NaCl2, start text, N, a, end text, plus, start text, C, l, end text, start subscript, 2, end subscript, right arrow, 2, start text, N, a, C, l, end text How many grams of \text{NaCl}NaClstart text, N, a, C, l, end text will be produced from 18.0 \text{ g}18.0 g18, point, 0, start text, space, g, end text of \text{Na}Nastart text, N, a, end text and 23.0 \text{ g}23.0 g23, point, 0, start text, space, g, end text of \text{Cl}_2Cl 2 ​ start text, C, l, end text, start subscript, 2, end subscript?

Answer 1

Answer: The mass of $NaCl$ produced is, 45.8 grams.

Explanation : Given,

Mass of $Na$ = 18.0 g

Mass of $Cl_2$ = 23.0 g

Molar mass of $Na$ = 23 g/mol

Molar mass of $Cl_2$ = 71 g/mol

Molar mass of $NaCl$ = 58.5 g/mol

First we have to calculate the moles of $Na$ and $Cl_2$.

$\text{Moles of }Na=\frac{\text{Given mass }Na}{\text{Molar mass }Na}$

$\text{Moles of }Na=\frac{18.0g}{23g/mol}=0.783mol$

and,

$\text{Moles of }Cl_2=\frac{\text{Given mass }Cl_2}{\text{Molar mass }Cl_2}$

$\text{Moles of }Cl_2=\frac{23.0g}{35.5g/mol}=0.648mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$2Na+Cl_2\rightarrow 2NaCl$

From the balanced reaction we conclude that

As, 2 mole of $Na$ react with 1 mole of $Cl_2$

So, 0.783 moles of $Na$ react with $\frac{0.783}{2}=0.392$ moles of $Cl_2$

From this we conclude that, $Cl_2$ is an excess reagent because the given moles are greater than the required moles and $Na$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NaCl$

From the reaction, we conclude that

As, 2 mole of $Na$ react to give 2 mole of $NaCl$

So, 0.783 mole of $Na$ react to give 0.783 mole of $NaCl$

Now we have to calculate the mass of $NaCl$

$\text{ Mass of }NaCl=\text{ Moles of }NaCl\times \text{ Molar mass of }NaCl$

$\text{ Mass of }NaCl=(0.783moles)\times (58.5g/mole)=45.8g$

Therefore, the mass of $NaCl$ produced is, 45.8 grams.