a student was asked to find the relative atomic mass of element X. crystals of the chloride of X were know to have the formular XCl2 .6H2O.the student dissolved 2.03g of the crystals in water, and then added an excess of silver nitrate solution to the solution formed. a white percipitate of silver chloride was formed, which was filtered, dried and weighed. 2.87g were formed. Ag+Cl---AgCl calculate: the number of moles of silver chloride formed? the number of moles of X chloride in the solution? the mass of 1 mole of XCl2.6H2O? the relative atomic mass of X?
Solution: XCl2*6H2O = XCl2 + 6H2O XCl2 + 2AgNO3 = 2AgCl↓ + X(NO3)2 n (moles) = m / M n (AgCl) = 2.87 g / 143.32 g/mol = 0.02 mol XCl2*6H2O = n (XCl2) = n (AgCl) / 2 = 0.02 / 2 = 0.01 mol m (H2O in compound) = 0.01 mol * 108 g/mol =1.08 g m (XCl2) = 2.03 g - 1.08 g = 0.95 g m (X) = 0.95 g - (0.01 mol * 71 g/mol) = 0.24 g Molar mass (X) = 0.24 / 0.01 =24 g/mol, so X is magnesium.