Question

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?

Answer 1

a student was asked to find the relative atomic mass of element X. crystals of the chloride of X were known to have the formula 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 precipitate 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?

$$XCl_2 + 2AgNO_3 \rightarrow 2AgCl + X(NO_3)_2$$ $$\nu(AgCl) = \frac{2.87}{143} = 0.02mol$$ $$\nu(XCl_2) = 0.01mol$$ $$\omega(XCl_2) = \frac{X + 71}{X + 71 + 18*6}$$ $$m(XCl_2) = 2.03 * \frac{X + 71}{X + 179}$$ $$\nu(XCl_2) = \frac{2.03}{X + 71} * \frac{X + 71}{X + 179} = 0.01mol$$ $$0.01X + 1.79 = 2.03$$ $$X = 24$$ $$M(XCl_2) = 96g/mol$$ $$M(X) = 24g/mol$$ $$X - Mg$$