A 0.50 g sample of a Group 2 metal, M, was added to 40.0 cm3 of 1.00 mol dm–3 hydrochloric acid (an excess). equation 1 M(s) + 2HCl (aq) → MCl 2(aq) + H2(g) (a) Calculate the amount, in moles, of hydrochloric acid present in 40.0 cm3 of 1.00 mol dm–3 HCl. (b) When the reaction had finished, the resulting solution was made up to 100 cm3 in a volumetric flask. A 10.0 cm3 sample of the solution from the volumetric flask required 15.0 cm3 of 0.050 mol dm–3 sodium carbonate solution, Na2CO3, for complete neutralisation of the remaining hydrochloric acid. (i) Write the equation for the complete reaction of sodium carbonate with hydrochloric acid. (ii) Calculate the amount, in moles, of sodium carbonate needed to react with the hydrochloric acid in the 10.0 cm3 sample from the volumetric flask. (iii) Calculate the amount, in moles, of hydrochloric acid in the 10.0 cm3 sample. (iv) Calculate the total amount, in moles, of hydrochloric acid remaining after the reaction shown in equation 1. (v) Use your answers to (a) and (b)(iv) to calculate the amount, in moles, of hydrochloric acid that reacted with the 0.50 g sample of M. (vi) Use your answer to (v) and equation 1 to calculate the amount, in moles, of M in the 0.50 g sample. (vii) Calculate the relative atomic mass, Ar, of M and identify M.
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