If you begin with 100.0 mL of a 3.0 Molar solution of ammonium sulfate and 100mL of a 2.0 Molar solution of barium phosphate how many grams of Barium sulfate would be produced?
To find the number of grams of barium sulfate produced, we need to determine the limiting reactant between the ammonium sulfate (NH4)2SO4 and the barium phosphate (Ba3(PO4)2).
Step 1: Calculate the number of moles for each compound.
- For ammonium sulfate:
- Volume = 100.0 mL
- Concentration = 3.0 Molarity
- Molar mass = 132.14 g/mol (NH4)2SO4
- For barium phosphate:
- Volume = 100 mL
- Concentration = 2.0 Molarity
- Molar mass = 601.93 g/mol Ba3(PO4)2
To find the number of moles, we use the formula:
Number of moles = Volume (in liters) x Concentration
For ammonium sulfate:
Number of moles = 100.0 mL x (1 L / 1000 mL) x 3.0 mol/L = 0.3 moles
For barium phosphate:
Number of moles = 100 mL x (1 L / 1000 mL) x 2.0 mol/L = 0.2 moles
Step 2: Find the limiting reactant.
The limiting reactant is the one with the smaller number of moles. In this case, barium phosphate (0.2 moles) is the limiting reactant.
Step 3: Determine the stoichiometry of the reaction.
The balanced chemical equation for the reaction between ammonium sulfate and barium phosphate is:
(NH4)2SO4 + Ba3(PO4)2 → 3(NH4)3PO4 + BaSO4
From the equation, we can see that 1 mole of barium phosphate produces 1 mole of barium sulfate (BaSO4).
Step 4: Calculate the number of moles of barium sulfate produced.
Since the number of moles of barium phosphate is equal to the number of moles of barium sulfate due to the stoichiometry of the reaction, the number of moles of barium sulfate produced is also 0.2 moles.
Step 5: Convert moles to grams.
To convert moles to grams, we need to multiply the number of moles by the molar mass of barium sulfate (BaSO4), which is 233.39 g/mol.
Number of grams of BaSO4 = 0.2 moles x 233.39 g/mol = 46.678 g
Rounded to three significant figures, the mass of barium sulfate produced is 46.7 grams.