How many grams of aluminum sulfate are produced if 15.7 g Al reacts with excess CuSO4?
3CuSO4 + 2Al ==> Al2(SO4)3 + 3Cu
mols Al = grams/atomic mass
Using the coefficients in the balanced equation, convert mols Al to mols Al(SO4)3 produced.
Then g Al2(SO4)3 = mols x molar mass = ?
To determine the number of grams of aluminum sulfate produced in the reaction, we need to first balance the chemical equation for the reaction between aluminum (Al) and copper sulfate (CuSO4). Once balanced, we can use stoichiometry to calculate the amount of aluminum sulfate formed.
The balanced chemical equation for the reaction between Al and CuSO4 is:
2 Al + 3 CuSO4 -> Al2(SO4)3 + 3 Cu
From the balanced equation, we can see that 2 moles of Al react with 3 moles of CuSO4 to produce 1 mole of Al2(SO4)3. Now, let's calculate the molar mass of Al2(SO4)3.
The molar mass of Al2(SO4)3 can be calculated as follows:
2(atomic mass of Al) + 3(atomic mass of S) + 12(atomic mass of O) = 2(26.98 g/mol) + 3(32.07 g/mol) + 12(16.00 g/mol) = 342.15 g/mol
Now, we can use stoichiometry to calculate the amount of Al2(SO4)3 produced.
Given: Mass of Al = 15.7 g
Molar mass of Al = 26.98 g/mol
Molar mass of Al2(SO4)3 = 342.15 g/mol
Step 1: Calculate the number of moles of Al using the given mass and molar mass:
Number of moles of Al = Mass of Al / Molar mass of Al = 15.7 g / 26.98 g/mol
Step 2: Use the balanced equation to relate moles of Al to moles of Al2(SO4)3:
2 moles of Al = 1 mole of Al2(SO4)3
Moles of Al2(SO4)3 = (Number of moles of Al) / 2
Step 3: Calculate the mass of Al2(SO4)3 using the moles obtained in step 2 and 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, you can calculate the number of grams of aluminum sulfate produced if 15.7 g of aluminum reacts with excess CuSO4.