Given:2 Na3PO4 + 3 CaCl2 ⟶ Ca3(PO4)2 + 6 NaCl844 grams of sodium phosphate and 891 grams of calcium chloride are supplied. If 719 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.

For Na3PO4: (233) + (31) + (164) = 164 g/mol

For CaCl2: (40) + (35.45*2) = 110.9 g/mol

For Ca3(PO4)2: (403) + (2((31) + (16*4))) = 310.18 g/mol

Step 2: Determine the theoretical yield.

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

For Na3PO4: 844 g / 164 g/mol = 5.15 mol

For CaCl2: 891 g / 110.9 g/mol = 8.03 mol

The balanced chemical equation tells us that it takes 2 moles of Na3PO4 to react with 3 moles of CaCl2 to produce 1 mole of Ca3(PO4)2.

So, the limiting reactant is Na3PO4 because 5.15 mol / 2 < 8.03 mol / 3.

The theoretical yield of Ca3(PO4)2 is therefore 5.15 mol / 2 * 310.18 g/mol = 799.3 g.

Step 3: Calculate the percent yield.

The percent yield is the actual yield (what was actually produced in the experiment) divided by the theoretical yield (what should have been produced, according to stoichiometry), multiplied by 100%.

Percent yield = (719 g / 799.3 g) * 100% = 89.95%

So, the percent yield is 89.95%.