Can someone help me solve these chemistry impure samples and percentage purity problems?

Question

Solving 1 or both of them is fine, as long as explanation is included.

1. An impure sample of Na2SO4 has a mass of 1.56 grams. This sample is dissolved and allowed to react with BaCl2 solution. The precipitate has a mass of 2.15g. Calculate the percentage of Na2SO4.

2. An impure sample of Na2SO4 has a mass of 1.65g and is dissolved in water. Addition of BaCl2 solution produced a precipitate of barium sulphate with mass 2.32g. What is the percentage of NaCl in the mixture?

Answer 1

1) Na2SO4 + BaCl2 --> BaSO4(s) + 2NaCl

1 mole of BaSO4 forms from 1 mole of 1 mole of Na2SO4
(2.15/233)0.009223 moles forms from.....????

0.009223X1
0.009223 moles
mass=moles X relative atomic mass
mass=0,009223 X 142
=1.31 g
percentage purity= mass of pure/ impure X 100

1.30/2.15 X100
=60.1 %

2) Na2SO4 + BaCl2 --> BaSO4(s) + 2NaCl

1 mole of BaSO4 formed from 1 mole of Na2SO4
(2.32/233) 0.01 moles formed from......??

0.01X1=0.01 moles

mass=0.01 X 142
=1.42 g

1 mole of BaSO4 also formed 2 moles of NaCl
0.01 moles formed ...??

0.01 X2= 0.02 moles

mass= 0.02 X 58.5
=1.17 g

% mass of NaCl= 1.17/(1.17+2.23+1.42 of impure Na2SO4) X100

=1.17/4.82 X100
=24.27 % of NaCl

Answer 2

Use moles. That is always the key to solving problems involving masses of different chemical substances.

You also need to know the reacting ratios, which in this case are obvious, but you may as well write the balanced equation to be on the safe side.

Na2SO4 + BaCl2 = BaSO4 + 2 NaCl

Some people like to use a roadmap:

g BaSO4 --> moles BaSO4 --> moles Na2SO4 --> g Na2SO4

2.15 g BaSO4 divided by formula mass of BaSO4 gives you moles BaSO4

From balance equation, each BaSO4 came from one Na2SO4.

So now you know how many moles Na2SO4 you had. Multiply by molar mass of Na2SO4, and that will give you g Na2SO4 in your sample.

In 2, I presume that you are meant to assume that whatever wasn't Na2SO4 in your sample was NaCl.

Over to you.

Answer 3

1. get the molar mass of na2so4, just add up the molar masses from the periodic tabl3.
2. once you have the molar mass, 1.56/molar mass to get the number of moles.
3. Get the equation for the reaction.
I think its like this:
Na2SO4 + BaCl2-----> 2NaCl + BaSO4, don't quote me, look it up, but i think its right.
The percentage of Na2SO2 is given by how much is left in the solution after rx, find the limiting reagent. If there is excess BaCl2, then the answer is zero. Otherwise you can't really tell.

Maybe that is not the right equation.....

Answer 4

At STP, one mole of gas takes up 22.4 liters. You only have 0.225 liters, so you'll have less than one mole. Calculate the number of moles you actually have. It's going to have a few zeros after that decimal! (1 mole)/(22.4 liters) x (0.225 liters) = 0.0xxx moles. For every mole of SO2, you needed TWO moles of H2SO4. So if you double your number, 0.0xxx moles x 2, that tells you how many moles of H2SO4 you started with. Now, calculate the molecular mass (aka formula mass, formula weight, molecular weight....) for H2SO4. That's how many grams are in ONE mole. It'll be somewhere around a hundred or so. Let's say it's 555 (it isn't!). That means that there are 555 grams for ONE mole. 555 g/mol. But you have (0.0xxx x 2) moles. Multiply the formula mass (aka 555 g/mol) by the number of moles, and you are done. (555)(0.0xx x2) = answer in grams.

Answer 5

go ahead and write your balaced chemical equation. thats the first step.