Question

Experiment 9 PRE-LABORATORY PROBLEMS 1. Calculate the molar masses of the following compounds a. strontium chloride hexahydrate, SrCl, 6H,O b. sodium phosphate dodecahydrate, Na, PO 12H,O and c. Strontium phosphate, Sr,(PO) 2. When a hydrated salt dissolves in water, what happens to the water in the salt? 3. What is the overall balanced reaction when an aqueous solution of SrCl, reacts with an aqueous solution of Na,P0,? 4. 2.000 g of Na, PO, 12H,O is reacted with 2.000 g of SrCl, 6H,O, calculate the theoretical yield of Sr,(PO), precipitate. Complete the following steps to solve this problem. a. Calculate the theoretical yield (mass) of Sr,(PO), produced from 2.000 g of Na,PO 12H,O if we assume that the strontium chloride salt is in excess. b. Calculate the theoretical yield of Sr,(PO,), produced from 2.000 g of SrCl, 6H,O again assuming that the sodium phosphate is in excess. c. Which reactant produces the smallest theoretical yield? d. Which compound above is the limiting reactant? EXPERIMENT Pg. 150

Answer 1

Solution 1. Calculate the a. strontium molar mass of the following compounds chloride hexahydrate, Srl, 6160 element Count Atom mass 87.62 35.453 H 12 15.99 molar mass = 8762 + 2 (35.453) +0x12) + (6x15.99 - 266.6148 g/mol b. Sodium phosphate dodecahydrate, Na, Poq ta tho Atom mass element Cound Na 22.98 3 30.97 Р 15.999 24 H 15.99 molar mass = (22.98 x 3) + (30.98 x 1 )+ (4 x 15-999) + (24x) + (2x18-19) = 380.117 ginol

c. phosphate, Srz PO4)2 Strontium element Count Atomic mass 3 87-62 30.974 15.999 Molar mass = (87.62 x3) + (30.944) + 2) + (15-999+8) 452.8 glmol When a hydrated salt dissolves in water, what happens to the water in the salt a d Solution Here the water in the sett hydrated salt Add cep to the other water moleceules. That is it combined with the large number of water. This is because when hydrated salt dissolves in water the cations are sumounded by water molecule with a negative oxygent and closer to positive Cation and the anions are surrounded by water molecules with the positive hydrogens closer to the negative anion thus they simply combine with large water molendes in the Solvat the solution. in the Solvent of

3. 2 Naz Pog. 12tho + 3 So Clubho - Sr, poq), to Naci+ 42420 Solution 48.a) theorectical gield = 9 Na Pon 12tho is the limiting reactent So balanced equation 2 Nagpoa. 12 ho + 3 So Clu6tho Srz (PO4)2 + 6 NaCl +424ho 2000 g of Naz poco 12tho = 22000 - 5.261 mol of Napos, 12H2O 380.117g 2000 g of Sx Clg Etho 2060 266.lt - 7.50 mol of SoCl2-6tho ... 5.261 Naz Poz. 1210 3 x 1 molg Stg (204)₂ 2 mol of Na Po 4 12 H₂O = 2.6305 moles : 45& 8gx 2.6305 = 1191-09 gumy This is the theorectical gield. Ime 04/2 7.50 So ClubHox 3 mol of Sochisho - 2.5 moles = 2.8 x 45 2.89 g 51004) = 1132 grams

4.6 Solution So (1.6h produced. Smallest theorectionl geild. 4d. Solution Stronium ora 20 is the limiting chloride reactant