Ba(NO3)2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaNO3(aq). The experiment was begun with 75.00g of Ba(NO3)2 . Calculate the theoretical yield of NaNO3.

Answer 1

 

Brenda M.

 asked •

Hey Brenda : ) ,

Here is the easiest way to arrive at our solution,

find the molar mass of Barium nitrate = 261.3 g/mol

find the molar mass of the desired product Sodium Nitrate = 85 g/mol

Use dimensional analysis, and don’t forget the mol to mol ratio from the balanced chem equation, to convert the moles of the reactant into the moles of the desired product. you got this! : )

Remember, in chemistry we usually start with the given quantity, in this case the 75g Barium Nitrate,

75g Ba(NO3)2 x [1mol Ba(NO3)2 / 261.3 g] x [2mol NaNO3 / 1mol Ba(NO3)2] x [85g / 1mol NaNO3] = 48.8g

 

48.79g Sodium Nitrate

Answer 2

Ba(NO₃)₂(aq) + Na₂SO₄(aq) ==> BaSO₄(s) + 2NaNO₃(aq) .. balance equation

Since we aren't given the amount of Na₂SO₄ that is used in this reaction, we must assume it is present in excess. Therefore, only the 75.00 g of Ba(NO₃)₂ will determine how much NaNO₃ can be formed.

First, we must convert the 75.00 g Ba(NO₃)₂ to moles:

molar mass Ba(NO₃)₂ = 261.3 g / mole

75.00 g Ba(NO₃)₂ x 1 mol / 261.3 g = 0.2870 mols

Next, we will use the stoichiometry of the balanced equation to find the theoretical yield (mols) of NaNO₃

0.2870 mols Ba(NO₃)₂ x 2 mols NaNO₃ / 1 mol Ba(NO₃)₂ = 0.5740 mols NaNO₃ = theoretical yield

This is the theoretical yield. But most times when asked for the theoretical yield, they want the answer in grams. So, we can now convert this answer to grams NaNO₃ using the molar mass of NaNO₃

molar mass NaNO₃ = 84.99 g / mol

0.5740 mols NaNO₃ x 84.99 g / mol = 48.78 g NaNO₃ (4 sig. figs.) = theoretical yield in grams