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)
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.
Calculate the total amount, in moles, of hydrochloric acid remaining after the reaction shown in equation 1.
amount = ... mol
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.
amount = ... mol
Use your answer to (v) and equation 1 to calculate the amount, in moles, of M in the 0.50 g sample.
amount = ... mol Calculate the relative atomic mass, Ar, of M and identify M.
Ar of M = ... identity of M = ...
The answer to your question is provided in the image:
Comments
Leave a comment