Aluminum sulfate reacts with barium chloride to yield barium sulfate and aluminum chloride. What mass of barium sulfate is produced from 650 mL of 0.320 M of aluminum sulfate?
Thank you.
Here is a sample problem I posted on stoichiometry. Just follow the procedure outlined. moles Al2(SO4)3 = M x L.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To find the mass of barium sulfate produced, we need to determine the moles of aluminum sulfate and then use the stoichiometry of the balanced equation to calculate the moles of barium sulfate. Finally, we can convert the moles of barium sulfate to grams using its molar mass.
First, let's calculate the moles of aluminum sulfate. We have the volume (650 mL) and the molarity (0.320 M). The formula to calculate moles is:
moles = volume (L) x molarity
Convert the volume to liters by dividing it by 1000:
650 mL ÷ 1000 = 0.650 L
Now we can calculate the moles of aluminum sulfate:
moles of aluminum sulfate = 0.650 L x 0.320 M = 0.208 mol
Next, we need to use the stoichiometry of the balanced equation to determine the mole ratio between aluminum sulfate and barium sulfate. From the balanced equation:
2Al₂(SO₄)₃ + 3BaCl₂ → 3BaSO₄ + 2AlCl₃
The mole ratio between aluminum sulfate and barium sulfate is 2:3.
To calculate the moles of barium sulfate formed, we can use the following equation:
moles of barium sulfate = (moles of aluminum sulfate) x (3 moles of barium sulfate / 2 moles of aluminum sulfate)
moles of barium sulfate = 0.208 mol x (3/2) = 0.312 mol
Finally, we can calculate the mass of barium sulfate by multiplying the moles by its molar mass. The molar mass of barium sulfate (BaSO₄) is:
Ba: 137.33 g/mol
S: 32.07 g/mol
O (4 atoms): 4 x 16.00 g/mol = 64.00 g/mol
Molar mass of BaSO₄ = 137.33 g/mol + 32.07 g/mol + 64.00 g/mol = 233.40 g/mol
mass of barium sulfate = moles of barium sulfate x molar mass of barium sulfate
mass of barium sulfate = 0.312 mol x 233.40 g/mol = 72.68 g
Therefore, approximately 72.68 grams of barium sulfate is produced from 650 mL of 0.320 M aluminum sulfate.