Calculate the mass of Al2(SO4)3 produced when 4.26kg of Al(NO3)3 is added to an excess of K2SO4.
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.
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.
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.
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
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.
To calculate the mass of Al2(SO4)3 produced, we need to determine the stoichiometry of the reaction between Al(NO3)3 and K2SO4, and then use the given mass of Al(NO3)3 to find the mass of Al2(SO4)3 produced.
The balanced equation for the reaction between Al(NO3)3 and K2SO4 can be written as:
2 Al(NO3)3 + 3 K2SO4 → Al2(SO4)3 + 6 KNO3
From the equation, we can see that 2 moles of Al(NO3)3 react to produce 1 mole of Al2(SO4)3. This means that the molar ratio between Al(NO3)3 and Al2(SO4)3 is 2:1.
To calculate the moles of Al(NO3)3 given the mass, we need to use the molar mass of Al(NO3)3. The molar mass of Al(NO3)3 is calculated by adding up the atomic masses of aluminum (Al), nitrogen (N), and three oxygen (O) atoms:
Molar mass of Al(NO3)3 = 1 * atomic mass of Al + 3 * atomic mass of N + 9 * atomic mass of O
Next, we calculate the moles of Al(NO3)3 by dividing the given mass by the molar mass of Al(NO3)3:
moles of Al(NO3)3 = mass of Al(NO3)3 / molar mass of Al(NO3)3
Once we have the moles of Al(NO3)3, we can use the stoichiometry to calculate the moles of Al2(SO4)3 produced. Given the 2:1 mole ratio between Al(NO3)3 and Al2(SO4)3, we can write:
moles of Al2(SO4)3 = (moles of Al(NO3)3) / 2
Finally, we can calculate the mass of Al2(SO4)3 by multiplying the moles by the molar mass of Al2(SO4)3:
mass of Al2(SO4)3 = moles of Al2(SO4)3 * molar mass of Al2(SO4)3
By following these steps, we can determine the mass of Al2(SO4)3 produced when 4.26 kg of Al(NO3)3 is added to an excess of K2SO4.