Question

A reaction is done by dissolving 5.687 g sodium phosphate and 7.533 g of calcium chloride in water. These react to form solid calcium phosphate and sodium chloride (this remains dissolved in water). a) Write a balanced equation for the reaction; don’t forget states of matter b) How many grams of calcium phosphate are expected to form if all the sodium phosphate reacts? c) How many grams of calcium phosphate are expected to form if all the calcium chloride reacts? d) Which is the limiting reagent and what is the theoretical yield of the calcium phosphate? e) Experimentally 4.312 g of solid was collected. What is the percent yield? f) How many grams of the excess reagent are leftover?

Answer 1

(a) 2 Na₃PO, (1) + 3 Calę (as) amol 3 mol Ca, (PO )+ (NaCl (12) 1 mol senola - moles of Napoy = 5.681 moles of Call = 7.533 = 0.0347 = 0.0678 110.98 as allo sodium phosphates reacts so, it will be lenitig reages & will also guide the amount of calcium phosphate caz (pouda = 310-18 g/mol. by unitan method - 2 mol of Mapou reacts with 3nd Call So, 0.0347 , » Cobby = 3X0-0347 = 0.0520s moda Smilarly, 2 mol Napou giver lind calcium phosphate So, 0.0347 will give 0.0347 = 0.01735 ml (c) In present case, all Call can't = 0.01735x310.18 be consumed, for that we need more Ngeon Theoretical so if, done, then, Cans (Powda = 260.0678) 0226 ml (6) (d) = 5.382g so, if 4.312 is experimental yield, % yield = experimental yield 100- Theoretical field" cach left excers = unrealted Ca C reacted with anal Ng PQ = 3 mol 0.0347 = 3x0.0347 = 0.052A5 reacted = 0.0678 0.05205 = 0 DIS15 ind = 001S 75 x 110.98 (4) = 1.748g Call yield 4.312 100= 80.117 5.382