Question

1.A 12.39g sample of phosphorus reacts with 45.4 g of Chlorine to form Phosphorus Trichloride(PCl3). if it PCL3 is the only product, what is the mass of pcl3 formed?

2.Determine the molarity of a solution formed by dissolving 8.47g LiBr in enough water to yield 750 mL of solution

3. Calculate the mass percent composition of Oxygen Al2(SO4)3

Answer 1

1)

2 P + 3 Cl₂ -------> 2 PCl₃

(12.39 g P) / (30.9738 g P/mol) = 0.400015 mol P
(42.54 g Cl2) / (70.9064 g Cl2/mol) = 0.599945 mol Cl2

0.400015 moles of P will react completely with 0.400015 x (3/2) = 0.600023 mol of Cl2, but there is not that much Cl2 present, so Cl2 is the limiting reactant.

Therefore,

weight of PCl3 formed is

(0.599945 mol Cl2) x (2/3) x (137.3334 g PCl3/mol) = 54.93 g PCl3

2) Molarity = (weight/molecular mass) x (volume in mL/1000) = 130.04 millimolar

3) Molecular Mass of Al2(SO4)3 is = 342.15

total number of oxygen = 12 Oxygens = 12 x 16 =192

Mass %age of oxygen = (192/342.15) x 100 = 56.1%