Question

19 grams of sodium and 34 grams of chlorine are measured out and put into separate glass vials that weigh 10 grams each. Then sodium is dumped into the glass vial containing the chlorine. They react and form a crystalline solid. What is the total weight of the glass vial containing the final product?

Answer 1

63g

Step-by-step explanation:

You don't have to make a lot of calculus to find the answer.

Just only remember the principle of the matter who says

"The matter is not created or destroyed only transformed"

It means, that even when you are transforming sodium and chlorine to obtain sodium chlorine, who is the crystalline solid, you can't obtain more matters (grams) that you put at the beginning of the experiment.

As well, you just have to add the grams of the sodium, chlorine and the recipient, glass vial containing the final product, as follow:

19 grams of sodium + 34 grams of chlorine + 10 grams glass vial

19 + 34 + 10 = 29 + 34 = 63 g

remember, the matter isn't create, just transformed, so, you can't add any more grams that you already have.

Answer 2

63.02 g.

Step-by-step explanation:

• Na reacts with Cl₂ according to the balanced equation:

2Na + Cl₂ → 2NaCl,

It is clear that 2 mole of Na react with 1 mole of Cl₂ to produce 2 moles of NaCl.

• Firstly, we need to calculate the no. of moles of Na and Cl₂:

no. of moles of Na = (mass/atomic mass) = (19.0 g/22.9897 g/mol) = 0.826 g.

no. of moles of Cl₂ = (mass/atomic mass) = (34.0 g/70.906 g/mol) = 0.48 g.

• From the stichiometry, Na reacts with Cl₂ with a molar ratio (2:1).

So, 0.826 mol of Na "the limiting reactant" reacts completely with 0.413 mol of Cl₂ "left over reactant".

The no. of moles of Cl₂ remained after the reaction = 0.48 mol - 0.413 mol = 0.067 mol.

∴ The mass of Cl₂ remained after the reaction = (no. of moles of Cl₂ remained after the reaction)(molar mass of Cl₂) = (0.067 mol)(70.906 g/mol) = 4.75 g.

• To get the no. of grams of produced NaCl:

using cross multiplication:

2 mol of Na produce → 2 mol of NaCl, from the stichiometry.

∴ 0.826 mol of Na produce → 0.826 mol of NaCl.

∴ The mass of NaCl produced after the reaction = (no. of moles of NaCl)(molar mass of NaCl) = (0.826 mol)(58.44 g/mol) = 48.27 g.

∴ The total weight of the glass vial containing the final product = the weight of the glass vial + the weight of the remaining Cl₂ + the weight of the produced NaCl = 10.0 g + 4.75 g + 48.27 g = 63.02 g.