How many grams of Al2(SO4)3 can be produced from 10.0g of CuSO4?
well, the bare-bones equation is of the form
Al2(SO4)3 + x -> 3CuSO4 + x
so, each mole of Al2(SO4)3 produces 3 moles of CuSO4.
Just convert your grams to moles, and work it from there.
Personally, I prefer the answer of "You can't".
DrBob222 may have the right of it.
And besides, my reaction "equation" is backwards. We wanted to produce the Al2(SO4)3, not use it.
But then, I'm sure you caught that, right?
To be honest about it I was just trying to be a smart alec! Turning CuSO4 into Al2(SO4)3 is like the alchemists who tried to turn Pb into Au.
What, there isn't any kind of double-replacement reaction that can swap metals like that?
Sure, but I read the problem as we had 10.0 g CuSO4 and nothing more; thus, the alchemist thought.
To determine the number of grams of Al2(SO4)3 that can be produced from 10.0g of CuSO4, we need to set up a balanced chemical equation and use stoichiometry.
The balanced chemical equation for the reaction between CuSO4 and Al is:
3CuSO4 + 2Al -> Al2(SO4)3 + 3Cu
From the equation, we can see that 3 moles of CuSO4 react with 2 moles of Al to produce 1 mole of Al2(SO4)3.
Now, let's calculate the molar masses of CuSO4 and Al2(SO4)3:
- The molar mass of CuSO4 is 159.61 g/mol (1 Cu atom + 1 S atom + 4 O atoms).
- The molar mass of Al2(SO4)3 is 342.15 g/mol (2 Al atoms + 3 S atoms + 12 O atoms).
Next, we will use stoichiometry to find the amount of Al2(SO4)3 produced:
- Convert the mass of CuSO4 to moles using its molar mass: 10.0g CuSO4 * (1 mol/159.61 g) = 0.0627 moles of CuSO4.
- Since the stoichiometric ratio of CuSO4 to Al2(SO4)3 is 3:1, we can calculate the moles of Al2(SO4)3 produced: 0.0627 moles of CuSO4 * (1 mol Al2(SO4)3/3 mol CuSO4) = 0.0209 moles of Al2(SO4)3.
- Finally, convert moles of Al2(SO4)3 to grams using its molar mass: 0.0209 moles * 342.15 g/mol = 7.15 grams of Al2(SO4)3.
Therefore, 10.0 grams of CuSO4 can produce approximately 7.15 grams of Al2(SO4)3.