Question

10Na + 2NaNo3 yields 6Na2O + N2

how many grams of sodium oxide will be made by combining 25.0 grams of sodium with 47.0 grams of sodium nitrate?

Answer 1

Here we are talking about a limiting reactant and a excess reactant. The limiting reactant limits the amount of product that can be form in the reaction, so the amount of product will be depending on the amount of this reactant, when the limiting reactant is consumed the reaction stop and certain amount of product will form, that is why this reactant will determine the reaction. The excess reactant is the one that remains when the reaction is complete, since the reaction will stop when the limiting reactant is consumed.

We are going to obtain the grams of sodium oxide calculating which reactant is the limiting reactant and which is the excess reactant. The less amount of reactant will determine the amount (grams) of sodium oxide obtained in the reaction.

We use stoichiometry converting grams of reactant into mols of reactant, then in mols of product and grams of product.

stoichiometry: grams of reactant -> mols of reactant -> mols of product -> grams of product

10 Na+ + 2 NaNO3 -------> 6 Na2O + N2

grams of Na used in the reaction = 25.0 g

grams of NaNO3 used in the reaction = 47.0 g

Molar Mass Na = 22.98976 g/mol

Molar Mass NaNO3 = 84.995 g/mol

Molar Mass Na2O = 61.979 g/mol

1. Calculate how much product is produced by each reactant:

Na => 25 g of Na (1 mol Na / 22.98976 g Na) = 1.087 mols of Na

          1.087 mols of Na ( 6 Na2O / 10 mols Na) = 0.652 mols of Na2O

          0.652 mols of Na2O ( 61.979 g Na2O / 1 mol Na2O) = 40.4 g Na2O

NaNO3 => 47 g NaNO3 (1 mol NaNO3 / 84.995 g NaNO3) = 0.553 mols of NaNO3

                0.553 mols of NaNO3 ( 6 mols Na2O / 2 mols NaNO3) = 1.659 mols of Na2O

               1.659 mols of Na2O ( 61.979 g Na2O / 1 mol Na2O ) = 102.8 g of Na2O

Amount of product obtained with sodium = 40.4 g

Amount of product obtained with sodium nitrate = 102.8 g

As we can see the limiting reactant is sodium (Na) since we obtain the less amount of sodium oxide (Na2O) from it. It means that the amount of product obtained in this reaction is 40.4 g of Na2O.