Question

0.800 g of a metal Y (RAM = 40) reacted with 40.0 cm3 of 1.00 mol dm-3 hydrochloric acid to produce a neutral solution of Y chloride and 480 cm3 of hydrogen gas (H₂) collected at room temperature and pressure. Evaporating the solution to dryness produced 2.22 g of solid Y chloride.

(a) Find the empirical formula for Y chloride.
(b) Use the given information to construct the equation for the reaction.
(Cl = 35.5. Molar volume of a gas at RTP = 24.0 dm3)

Answer 1

The empirical formula of Y chloride, based on the given masses and the molar volume of a gas at room temperature and pressure (RTP), is YCl2. The balanced equation for the reaction is Y + 2HCl → YCl2 + H2.

Step-by-step explanation:

To find the empirical formula for Y chloride, we use the mass of the metal (0.800 g) with a relative atomic mass (RAM) of 40. This gives us a molar quantity of 0.020 mol of Y. The mass of Y chloride produced after evaporation is 2.22 g. Given that the RAM of Cl is 35.5, we can calculate the amount of Cl in the compound.

Since the remaining mass would be that of metal Y (2.22 g - 0.800 g = 1.42 g), we can then determine the moles of Cl in the compound by dividing by the RAM of Cl: 1.42 g / 35.5 g/mol = 0.040 mol. Therefore, the ratio of moles of Y to Cl is 1:2, resulting in the empirical formula YCl2.

To construct the equation for the reaction, we know that Y reacts with HCl to produce Y chloride and H2. Given the volume of H2 produced (480 cm3), we can calculate the moles of H2: 480 cm3 / 24000 cm3/mol = 0.020 mol. Since the ratio of H2 to Y is 1:1, the balanced equation is Y + 2HCl → YCl2 + H2.