4.50 grams of sodium phosphate is mixed with 3.75 grams of barium nitrate. How many grams of barium phosphate can be produced?
This is a limiting reagent (LR) problem. I work these the long way because I think this way is easier to explain.
Step 1. Write and balance the equation.
2Na3PO4 + 3Ba(NO3)2 ==> Ba3(PO4)2 + 6NaNO3
Step 2. Convert grams to mols.
mols Na3PO4 = grams/molar mass = ?
mols Ba(NO3)2 = grams/molar mass = ?
Step 3. Convert each mol of reactant from step 2 to mols of the product using the coefficients in the balanced equation.
a. mols Ba3(PO4)2 from Na3PO4 = mols Na3PO4 from step 2 x (1 mol Ba3(PO4)2/2 mols Na3PO4) = ?
b. mols Ba3(PO4)2 from Ba(NO3)2 = mols Ba(NO3)2 x (1 mol Ba3(PO4)2/3 mol Ba(NO3)2) = ?
c. It is likely that the value of mols Ba3(PO4)2 from a and b will not agree; the correct value in limiting reagent problems is ALWAYS the smaller value.
Step 4. Using the smaller value, convert mols Ba3(PO4)2 to grams.
g = mols x molar mass.
1.64 grams
To determine the number of grams of barium phosphate that can be produced, you need to figure out which reactant limits the production of barium phosphate. This is determined by the stoichiometry of the balanced chemical equation.
The balanced chemical equation for the reaction between sodium phosphate (Na3PO4) and barium nitrate (Ba(NO3)2) is:
3 Na3PO4 + 2 Ba(NO3)2 → Ba3(PO4)2 + 6 NaNO3
According to the equation, you need 3 moles of sodium phosphate to react with 2 moles of barium nitrate to produce 1 mole of barium phosphate.
Step 1: Convert the given masses of sodium phosphate and barium nitrate to moles.
- For sodium phosphate: Use the molar mass of Na3PO4 to convert grams to moles.
Molar mass of Na3PO4 = 3(22.99 g/mol) + 1(30.97 g/mol) + 4(16.00 g/mol) = 163.94 g/mol
Moles of sodium phosphate = 4.50 g / 163.94 g/mol
- For barium nitrate: Use the molar mass of Ba(NO3)2 to convert grams to moles.
Molar mass of Ba(NO3)2 = 137.33 g/mol + 2(14.01 g/mol) + 6(16.00 g/mol) = 261.34 g/mol
Moles of barium nitrate = 3.75 g / 261.34 g/mol
Step 2: Determine the limiting reactant.
The limiting reactant is the reactant that is completely consumed in the reaction and restricts the amount of product formed. To find the limiting reactant, compare the moles of each reactant to their stoichiometric coefficients in the balanced equation.
According to the balanced equation, the mole ratio of Na3PO4 to Ba(NO3)2 is 3:2. This means that the stoichiometric ratio between the two reactants is 3 moles of Na3PO4 to 2 moles of Ba(NO3)2.
To compare the moles, divide the moles of each reactant by their respective coefficients in the balanced equation.
- Sodium phosphate: Moles / coefficient = 4.50 g / 163.94 g/mol / 3 = X moles
- Barium nitrate: Moles / coefficient = 3.75 g / 261.34 g/mol / 2 = Y moles
The smaller value, X or Y, corresponds to the limiting reactant. So, the limiting reactant is the one that gives the smaller value.
Step 3: Calculate the moles of barium phosphate that can be produced using the limiting reactant.
Since the limiting reactant is barium nitrate, we can calculate the moles of barium phosphate produced using the mole ratio from the balanced equation.
According to the balanced equation, the mole ratio of Ba(NO3)2 to Ba3(PO4)2 is 2:1. This means that 2 moles of Ba(NO3)2 will produce 1 mole of Ba3(PO4)2.
So, the moles of Ba3(PO4)2 produced = Y moles (moles of barium nitrate) / 2
Step 4: Calculate the mass of barium phosphate produced using the moles and molar mass of Ba3(PO4)2.
Use the molar mass of Ba3(PO4)2 to convert moles to grams.
- Molar mass of Ba3(PO4)2 = 3(137.33 g/mol) + 2(31.00 g/mol) + 8(16.00 g/mol) = 601.93 g/mol
- Mass of barium phosphate = Y moles (moles of barium nitrate) / 2 × 601.93 g/mol
By following these steps, you should be able to determine the number of grams of barium phosphate that can be produced.