a 0.5000-g sample containing NaHCO3, Na2CO3, and H2O was dissolved and diluted t

Question

a 0.5000-g sample containing NaHCO3, Na2CO3, and H2O was dissolved and diluted to 250.0 mL. A 25.00-mL aliquot was then boiled with 50.00 mL of 0.01255 M HCl. After cooling, the excess acid in the solution required 2.34 mL of 0.01063 M NaOH when titrated to a phenolphthalein end point. A second 25.00-mL aliquot was then treated with an excess of BaCl2 and 25.00 mL of the base; precipitation of all the carbonate resulted, and 7.63 mL of the HCl was required to titrate the excess base. Calculate the composition of the mixture.

Explanation / Answer

First write balance equations for the reactions: Cu2O + H2 ---> 2 Cu + H2O CuO + H2 ---> Cu + H2O mol of pure Cu made = 1.293 g /63.54 = 0.0203 mol if x = number of g of Cu2O present then you can write the following equation based upon the reactions and molecular weight relationships Term #1 --- 2*(x/143.09) - where the 2 comes from the fact that the Cu2O gives 2 mol of Cu for every mol of Cu2O consumed, and the 143.09 is the MW of Cu2O Term # 2 --- ((1.514-x)/79.55) - where the 1.514 - x is the remaining mass of CuO and 79.55 is the MW of CuO.

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