8. A student weighed 0.2 g of unknown oxide into a 250 ml conical flask. 50.00 m

Question

8. A student weighed 0.2 g of unknown oxide into a 250 ml conical flask. 50.00 ml HCI was added immediately to the flask which reacted with the oxide. The excess HCI was then titrated with 0.05M NaOH (ag) Use the table and the equations below to answer the questions that follows MO +2H (10pts)Complete the table below Burctte Reading (ml) nitial Final Titre a 0.00 21.50 42 42.00 2100 20.5 22.5 b. (Spts)Calculate the average titre c. (1Spts)Ca)culate the moles of the NaOH that reacted in the titration olume d. (1Spts)Determine the moles of HCI that reacted with the base, e. (20pts)Calculate the initial moles of C1 before the titration and hence the moles HCI hat reactod with the oxide

Explanation / Answer

c) the number of moles of HCl titrated must be equal to the number of moles of NaOH titrated, I understand that the excess of HCl was titrated with NaOH (NaOH in the buret)

I will use the average titre volume 21.5 ml or 0.0215 L

moles of NaOH used = Molarity * Volume = 0.05 * 0.0215 = 0.001075 moles

d) The moles of HCl that reacted with the base are the same moles calculated in the last step, this is because reactino is 1 : 1, 1 mole of HCl reacts with only 1 mole of NaOH

moles of HCl reacted are 0.001075 moles

e) We need the initial concentration of HCl, you wrote that the Molarity of HCl is 0.1 M , volume is 50 ml or 0.05 L

Initial moles of HCl = 0.1 * 0.05 = 0.005 original moles

Moles that reacted with the oxide = Original moles - moles consumed in the titration = 0.005 - 0.001075 = 0.003925 moles of HCl that reacted with the oxide

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