Prelaboratory question help

1.       The complete ionic equation contains all the ions in solution even spectator ions.

If you look in the introduction they give you the net ionic equation which only contains what is involved in the precipitate.

SO42-(aq)   +  Ba2+    BaSO4(s)

The complete ionic would contain the ions that remain in the supernatant.

SO42-(aq) + K+ (aq)  +  Ba2+(aq)  +  Cl-(aq)  BaSO4(s)  Cl-(aq) +  K+ (aq)

I will let you balance the equation.

2.       To answer number two why would you need excess barium ions in the precipitation.

A hint would be to look at the purpose of the lab, determine the percentage of sulfate ions in an unknown mixture. So if the barium ions are limited what is the result in the products

3.       Using stoichiometry we can calcululate the mass of the sulfate in the unknown by using the mass of the precipitate.

SO42-(aq)   +  Ba2+    BaSO4(s)

?g                                   0.0676g

0.0676 g BaSO4  ÷  molar mass of barium sulfate x mole ratio x  molar mass of sulfate ions =

Grams of sulfate ion

The mass percent =  grams of ion (sulfate) / mass of sample (0.100g)

4.       Similar to an empirical formula problem from earlier but trying to find the ratio of water attached to Calcium sulfate. CaSO4 · ?H2O

Convert percent to grams by making the sample out of 100 grams so 55.8% sulfate ion becomes 55.8g, than convert to moles. So if you have X number of moles of sulfate ions than you should have same number of moles of calcium ion and working backward find the number of grams of calcium. Know you have both the mass of sulfate and the mass of Calcium. Add the two and you get the mass of calcium sulfate. Remember that the total mass is 100 grams so now subtract the mass of calcium sulfate by the 100 grams and you get the mass of water attached the calcium sulfate. Which you can convert moles and then compare to the moles of calcium sulfate and find the ratio.