Final answer:
To determine the grams of aluminum nitrate formed, we need to use the balanced equation and convert the volume of aluminum sulfate to moles. Then, using the molar mass of aluminum nitrate, we can calculate the mass formed.
Step-by-step explanation:
In order to determine how many grams of aluminum nitrate can form when 2000 ml of 0.500 M aluminum sulfate reacts with excess barium nitrate, we need to use the balanced chemical equation:
Al2(SO4)3 + 6Ba(NO3)2 -> 2Al(NO3)3 + 3Ba(SO4)
From the equation, we can see that 1 mole of aluminum sulfate (Al2(SO4)3) reacts with 6 moles of barium nitrate (Ba(NO3)2) to form 2 moles of aluminum nitrate (Al(NO3)3) and 3 moles of barium sulfate (Ba(SO4)).
To find the number of moles of aluminum nitrate formed, we need to convert the volume of aluminum sulfate solution to moles using the given concentration (0.500 M):
Moles of aluminum sulfate = concentration x volume = 0.500 M x 2.000 L = 1.000 mol
Since the reaction is in a 1:2 ratio, the number of moles of aluminum nitrate formed will be half of the moles of aluminum sulfate: 1.000 mol / 2 = 0.500 mol.
Now, to find the mass of aluminum nitrate formed, we can use its molar mass.
The molar mass of aluminum nitrate (Al(NO3)3) is:
Molar Mass = (molar mass of Al) + 3 x (molar mass of N) + 9 x (molar mass of O) = 26.98 g/mol + 3 x 14.01 g/mol + 9 x 16.00 g/mol = 213.00 g/mol.
Mass of aluminum nitrate = moles x molar mass = 0.500 mol x 213.00 g/mol = 106.50 g