Stoichiometry?

Question

What mass in grams of Na is needed to synthesize 5.05 grams of Na2S from its elements ?

Answer 1

Hey ^_^

well to answer you question, you first need to write out the chemical equation, so it would look soemthing like this:

2Na(s) + S(s) --> Na2S(s)

You know already that to find the mass of something you need to use the "mole" equation" given: n = m/M

Since you are given the mass of Na2S(s), you need to work backwards to obtain it's mole "value" then substitute that mole value into the mole equation to find the mass of Na. Here's how it looks:

m of Na2S = 5.05g
M of Na2S = 2Na + S
= 2(22.99g/mol) + 32.06 g/mol
= 78.04 g/mol
n= m/M
= 5.05g/ 78.04g/mol
= 0.0647mol

Now you need to find the mole of Na. You know that the Mole ratio of Na: Na2S is 2:1 [ referred to the balanced chemical equation] you need to find out the mole ratio by substituting the mole of Na2S:

n of Na/ n of Na2S = 2mol/1mol

Therefore:

n of Na = n of Na2S(2mol/1mol)
= 0.0647mol(2mol/1mol)
= 0.1294mol

Okay now to find your mass of Na, rearrange equation:

m = nM
= 0.1294mol x 22.99g/mol
= 2.97 g

Therefore, you are required to have about 2.97g of Na in order to syntheszie 5.05 g of Na2S from its elements.

Hope that helps! =]

Answer 2

Atomic weights: Na = 23 S = 32 Na2S = 78

2Na + S ===> Na2S

5.05gNa2S x 1molNa2S/78gNa2S x 2molNa/1molNa2S x 23gNa/1molNa =

The 5.05gNa is given. The first factor comes from the molecular weight. The g Na2S cancel, leaving moles Na2S. The second factor comes from the balanced equation. The moles Na2S cancel, leaving moles Na. The third factor comes from the atomic weight of sodium. The mol Na cancel, leaving grams Na, which is what was wanted.