Question

What is the theoretical yield of sodium chloride for the reaction of 36.0 g Na with 75.2 g Cl2 of the following 2NA+Cl2---> 2NaCl? I keep getting stuck with my conversions. I started with 36.0g NA (1 mol Na/22.99g Na) (1 mol Cl2/2mol Na) =0.782 mol Cl2.

Answer 1

The theoretical yield is 91.5g of `NaCl`

Method 1

Start with the equation:

`2Na_((s))+Cl_(2(g))rarr2NaCl_((s))`

So 2 mole `Na` reacts with 1 mole `Cl_2` to give 2 moles`NaCl`

Convert moles to grams: `A_r Na=23` `A_rCl=35.5`

2x23 = 46g `Na` reacts with 2x 35.5 = 71g `Cl_2` to give 2x(23 +35.5 ) = 117g `NaCl`

Use simple proportion to find how much would be made from 36g `Na` by first working out what 1 g of `Na` would give by dividing through by 46:

(46/46)g `Na` reacts with (71/46) g `Cl_2` to give (117/46)g `NaCl`

So:

1g `Na` reacts with 1.54g `Cl_2` to give 2.54g `NaCl`

So:

36g `Na` reacts with 1.54x36= 55.44g `Cl_2` to give 2.54x36=91.44g `NaCl`

You can see that the chlorine is in excess so the theoretical yield is 91.5g of `NaCl` to 1sf.

Method 2

Start with the balanced equation as above. Calculate the amount of NaCl that can form from each reactant.

36.0 g Na × `"1 mol Na"/"22.99 g Na" × "2 mol NaCl"/"2 mol Na"` = 1.566 mol NaCl (3 significant figures + 1 guard digit)

75.2 g Cl₂ × `("1 mol Cl"_2)/("70.91 g Cl"_2) × "2 mol NaCl"/("1 mol Cl"_2)` = 2.121 mol NaCl

The Na gives fewer moles of NaCl, so Na is the limiting reactant.

The theoretical yield is

1.566 mol NaCl × `"58.44 g NaCl"/"1 mol NaCl" =" 91.5 g NaCl"`

Method 3

Divide the moles of each reactant by its coefficient in the balanced equation.

36.0 g Na × `"1 mol Na"/"22.99 g Na" =" 1.566 mol Na"` (3 significant figures + 1 guard digit)

75.2 g Cl₂ × `("1 mol Cl"_2)/("70.91 g Cl"_2) =" 2.121 mol Cl"_2`

`2"Na" + "Cl"_2 → 2"NaCl"`

`1.566/2`; `2.121/1`

0.7829; 2.121

NaCl gives fewer "moles of reaction", so NaCl is the limiting reactant.

So theoretical yield is

0.7829 mol reaction × `"2 mol NaCl"/"1 mol reaction" × "58.44 g NaCl"/"1 mol NaCl"` = 91.5 g NaCl