2 Na3PO4 + 3 CaCl2 Ca3(PO4)2 + 6 NaCl825 grams of sodium phosphate and 890 grams of calcium chloride are supplied. If 737 grams of calcium phosphate are produced, compute for the % yield.Write your final answer in two decimal places.Use the following mass numbers:Na - 23P - 31O - 16Ca - 40Cl - 35.45

To solve this problem, we need to follow these steps:

1. Calculate the molar mass of each compound.
2. Determine the theoretical yield.
3. Calculate the percent yield.

Step 1: Calculate the molar mass of each compound.

• Sodium phosphate (Na3PO4): (323) + 31 + (416) = 164 g/mol
• Calcium chloride (CaCl2): 40 + (2*35.45) = 110.9 g/mol
• Calcium phosphate (Ca3(PO4)2): (340) + (2(31+(4*16))) = 310 g/mol

Step 2: Determine the theoretical yield.

First, we need to find out how many moles of each reactant we have:

• Moles of sodium phosphate = 825 g / 164 g/mol = 5.03 mol
• Moles of calcium chloride = 890 g / 110.9 g/mol = 8.02 mol

From the balanced chemical equation, we know that the reaction consumes sodium phosphate and calcium chloride in a 2:3 ratio. So, we need to find out which is the limiting reactant:

• Sodium phosphate can react with 5.03 mol / 2 = 2.515 mol of calcium chloride
• Calcium chloride can react with 8.02 mol / 3 = 2.673 mol of sodium phosphate

Since sodium phosphate can react with fewer moles of calcium chloride, it is the limiting reactant. Therefore, the theoretical yield of calcium phosphate is based on the moles of sodium phosphate:

• Theoretical yield = 5.03 mol * (1 mol Ca3(PO4)2 / 2 mol Na3PO4) = 2.515 mol
• Convert this to grams: 2.515 mol * 310 g/mol = 779.65 g

Step 3: Calculate the percent yield.

• Percent yield = (actual yield / theoretical yield) * 100%
• Percent yield = (737 g / 779.65 g) * 100% = 94.53%

So, the percent yield of the reaction is 94.53%.