Calculate the concentration mol dm3 and g dm 3 of Sodium ethanedioate (Na2C2O4) Solution 5.00cm3 of which were oxidized in an acid Solution by 24.50cm3 of a potassium manganate (vill solution containing 0.05mol (dm3​

Respuesta :

The mass concentration of Sodium ethanedioate is 82 g/dm^3 while its molar concentration is 0.6125 M.

What is a redox reaction?

The term redox reaction refers to a reaction that involves the loss/gain of electrons. The redox reaction here is shown as;

2MnO4^-(aq) + 5C2O4^2- (aq) + 16H^+ ---->2Mn^2+(aq) + 10CO2(g) + 8H2O(l)

Number of moles of permanganate = 0.05 * 24.50/1000 = 1.225 * 10^-3 moles

2 moles of permanganate reacts with 5 moles of ethanedioate

1.225 * 10^-3 moles of permanganate reacts with 1.225 * 10^-3 moles *  5 moles /2 moles = 3.1  * 10^-3 moles

Since concentration = number of moles / volume

concentration of ethanedioate = 3.1  * 10^-3 moles/ 5.00 * 10^-3

= 0.6125 M

Mass concentration = molar concentration * molar mass = 0.6125 M * 134 g/mol = 82 g/dm^3

Learn more about concentration: https://brainly.com/question/492238

ACCESS MORE
EDU ACCESS
Universidad de Mexico