Question

Barium nitrate (Ba(NO3)2) reacts with sodium chloride (NaCl) in a double replacement (displacement) reaction, shown below.

Ba(NO3)2(aq)+NaCl(aq)-->???

How many grams of barium salt are produced when a solution containing 21.7 g of Barium nitrate is mixed with a solution containing excess sodium chloride?

Use 261.34 as the molar mass for barium nitrate. Round to three significant digits.

Answer 1

you know that they will be a displacement reaction that will form a barium salt:

Ba(NO3)2+ 2NaCl--> BaCl2 + 2NaNO3

So now that we have that formula and the molecular weight we can determine how much salt will be made. So here we convert the grams to moles

(42.3g Ba(NO3)2)*(1 mole/261.34g) = 0.16185 mol

In the molecular formula we know that 1 mole of Barium nitrate will create 1 mole of Barium chloride, so in this case (in a perfect world) you should get 0.16185 mole of barium chloride (208.23 g/mol) that we then have to convert to grams.

(0.16185 mol BaCl2) * ( 208.23 g/mol) = 33.7037 g of Barium Chloride (rounded to 3 significant digits = 33.7g)