Ores are analyzed by dissolving some of the ore and then performing an analysis of a metal (expressed in grams); usually the concentration of the metal is expressed as % of the amount of the ore (in grams) dissolved. An ore containing V was dissolved and then titrated with MnO4-. When 5.1650 g was dissolved and titrated with 0.1467 M MnO4- the titration required 22.57 mL to reach the endpoint. What was the % V in the ore?
5VO2+(aq) + MnO4-(aq) + 11H2O(l) ¡ê 5V(OH)4+(aq) + Mn2+(aq) + 2H+(aq)
To find the percent V in the ore, we need to calculate the amount of V present in the 5.1650 g of ore that was dissolved and titrated.
First, let's determine the number of moles of MnO4- used in the titration. We can use the formula:
moles MnO4- = Molarity (MnO4-) * volume (in liters)
Given:
Molarity (MnO4-) = 0.1467 M
Volume = 22.57 mL = 22.57 / 1000 = 0.02257 L
moles MnO4- = 0.1467 * 0.02257 = 0.0033070 moles
Using the balanced equation, we can determine the stoichiometry between the MnO4- and V:
1 mole MnO4- : 5 moles V
Therefore, 0.0033070 moles of MnO4- react with (0.0033070 * 5) moles of V.
Now, let's calculate the molar mass of V based on the equation:
5VO2+ : 5 * molar mass of V + 2 * molar mass of O = 1 * molar mass of V : 259.0178 g/mol
Since there are (0.0033070 * 5) moles of V:
Mass of V = (0.0033070 * 5) * molar mass of V = 0.0165355 * 259.0178 = 4.2792 g
Therefore, the % V in the ore is calculated as:
Percent V = (mass of V / mass of ore) * 100
Percent V = (4.2792 / 5.1650) * 100 = 82.97%
The percent V in the ore is approximately 82.97%.