When heated, sodium metal combines with chlorine gas to form solid sodium chloride...?

Question

When heated, sodium metal combines with chlorine gas to form solid sodium chloride:

2NA + Cl2 --> 2NaCl

If 8.30 g of sodium is heated with an excess of chlorine, a total of 19.5 g of sodium chloride is produced. Determine the theoretical yield and percentage yield.

I missed the lesson at school, so im really confused. Can anyone help me please? Thanks.

Answer 1

So first you must convert the amount of grams of sodium which you are given into moles of sodium by dividing by the molecular weight of sodium (22.99) since ALL stoichiometry is done on a molar basis (as that's the way balanced equations are written). So, i'll do that below

8.30 g of Na/22.99 g/mol = 0.361 moles

then you look at the balance equations, you see that for each 2 moles of sodium which you consume, you produce 2 moles of sodium chloride (a 1:1 molar ratio). So now, we convert the moles of Sodium into moles of sodium chloride using this relationship

0.361 moles of Na * (2 mols NaCl/2 mol Na) = 0.361 moles of NaCl

That's the theoretical molar yield of NaCl but we want to know how many GRAMS we're supposed to get, so we do that by multiplying the number of moles by the molecular weight as shown below:

0.361 moles of NaCl * (58.44 g/mol) = 21.1 g of NaCl

So that's what we're supposed to get, to calculate percent yield, we use the following formula:

% yield = (actual yield/theoretical yield) * 100

since we actually made 19.5 grams of sodium chloride, our percent yield is calculated as follows:

% yield = 19.5/21.1 *100 = 92.4%

Answer 2

A simple stoich problem. You just need to remember to always change grams given to moles and work from there.

step 1
Change grams of limiting chemical to moles.

8.30 g Na / 23 g Na per mole Na from the periodic table = 0.357 mol Na used.

step 2
look at the mole ratio of reactants and products in your balanced equation.

2 Na reactants gives you 2 NaCl products, or 2 mol Na to 2 mol NaCl. That means that 0.357 mol Na should give you 0.357 mol NaCl.

step 3
change moles back to grams of products.

0.357 mol NaCl x 58.5 g NaCl per mol = 20.9 g NaCl

**The theoretical yield is 20.9 g NaCl**

Percent yield = (actual yield / theoretical yield) x 100

actual yield is given as 19.5 g NaCl
theoretical yield is calculated to be 20.9 g NaCl

19.5 / 20.9 = 0.933 x 100 = 93.3% yield

**Percent yield is 93.3%**