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?

63.02 g.

Explanation:

2Na + Cl₂ → 2NaCl,

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

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.

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.

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.

a c d and f

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