Question

How do I do this limiting reactants problem about barium sulfate?

`"BaCl"_2(aq) + "Na"_2"SO"_4(aq) -> "BaSO"_4(s) + 2"NaCl"(aq)`

Object
Mass
Flask
89.4 grams
Mass of BaSO4 + flask
252.44 grams
2nd weighing: Mass of BaSO4 + flask
248.33 grams

1.) What mass of BaSO4 did the students actually collect? How did you find this?

2.) The students went in the lab with 150 grams of BaCl and 100 grams of Na2SO4. Which of these reactants is the limiting reactant?

3.) How much BaSO4 should the students have found in the lab?

4.) What was the students percent error?

Answer 1

The reaction is given to you:

`"BaCl"_2(aq) + "Na"_2"SO"_4(aq) -> "BaSO"_4(s) + 2"NaCl"(aq)`

All these questions ask is:

1. Find the mass of the product.
2. Given these starting masses of reactants, which one runs out first?
3. What should the theoretical recovered mass of the product be?
4. What was the percent error compared to the theoretical value?

If you REALLY understand the process, you should be able to do the other three questions you had posted...

---

And so,

`1)` The mass of the product is found by subtraction. I assume the "second weighing" is from a second trial. If not, you'll have to specify...

`"252.44 g" - "89.4 g vessel" = "163.0 g first weighing"`

`"248.33 g" - "89.4 g vessel" = "158.9 g second weighing"`

`("163.0 g" + "158.9 g")/2 = color(blue)"161.0 g"` average mass

`2)` If the students know that they have `"150 g BaCl"_2` and `"100 g Na"_2"SO"_4`, then we find from the mole ratio which one is in excess.

`150 cancel("g BaCl"_2) xx ("1 mol BaCl"_2)/(137.327 + 2cdot 35.453 cancel("g BaCl"_2)) = "0.7203 mols BaCl"_2`

`100 cancel("g Na"_2"SO"_4) xx ("1 mol Na"_2"SO"_4)/(2 cdot 22.989 + 32.065 + 4cdot 15.999 cancel("g Na"_2"SO"_4)) = "0.7040 mols Na"_2"SO"_4`

The reaction calls for `"1 mol"` of each, so since there are less than `"0.7203 mols"` of `"Na"_2"SO"_4`, that is the limiting reactant, and `"0.0163 mols"` of `"BaCl"_2` remains in excess.

`3)` The theoretical yield is then the mass that SHOULD have been recovered in an ideal world... and it is found from the limiting reactant mass.

`color(blue)("Theoretical Yield") = 0.7040 cancel("mols Na"_2"SO"_4) xx cancel("1 mol BaSO"_4)/cancel("1 mol Na"_2"SO"_4) xx (137.327 + 32.065 + 4 cdot 15.999 cancel("g BaSO"_4))/cancel("1 mol BaSO"_4)`

`=` `color(blue)("164.3 g BaSO"_4(s))`

`4)` And so if the student recovered `"161.0 g"` on average, then

`color(blue)(%"Error") = ("Actual - Theoretical")/("Theoretical") xx 100%`

`= |(161.0 - 164.3)/(164.3)| xx 100%`

`= color(blue)(2.02%)` under the theoretical value

And the percent yield is `color(blue)(97.98%)`.