Calculate the mass of Al2(SO4)3 produced when 4.26kg of Al(NO3)3 is added to an excess of K2SO4.

Answer 1

Balance the reaction:

2 Al(NO3)3 + 3 K2SO4 ---> Al2(SO4)3 + 6 K2SO4

Convert mass of Al(NO3)3 to moles:

4.26kg of Al(NO3)3 = 4.26 x 10^3 g of Al(NO3)3

4.26 x 10^3 g of Al(NO3)3 *(1 mole/212.996g)= moles of Al(NO3)3

moles of Al(NO3)3*(1 mole of Al2(SO4)3/2 moles of Al(NO3)3)= moles of Al2(SO4)3

moles of Al2(SO4)3*(342.151g/1 mole of Al2(SO4)3= mass of Al2(SO4)3

Answer contains only three significant figures.

Answer 2

I don't think any Al2(SO4)3 will be produced because there is nothing to drive the reaction. The reasons for a reaction are
1. gas is formed. none here
2. ppt formed. everything soluble here.
3. Slightly ionized material. not true here.

But in the spirit of the question this is how to calculate it.
1. Write and balance the equation.
2Al(NO3)3 + 3K2SO4 ==> Al2(SO4)3 + 6KNO3
2.mols Al(NO3)3 = grams/molar mass
3. Using the coefficients in the balanced equation, convert mols Al(NO3)3 to mols Al2(SO4)3.
4. Convert mols Al2(SO4)3 to g. g = mols x molar mass.

Answer 3

I didn't think that this reaction would occur, but they way that I went about doing it is how you would solve it. The reaction, balanced is:

2 Al(NO3)3 + 3 K2SO4 ---> Al2(SO4)3 + 6 KSO4

and not

2 Al(NO3)3 + 3 K2SO4 ---> Al2(SO4)3 + 6 K2SO4

I had a typo.

Answer 4

The procedure I wrote and the one Devron wrote is essentially the same. Assuming the reaction proceeds, the correct balanced equation is
2Al(NO3)3 + 3K2SO4 ==> Al2(SO4)3 + 6KNO3

Answer 5

Dr. Bob222 is correct: the setups are more or less the same. I just corrected my typo so that you did not get confused looking at what he wrote and looking at what I wrote.