The vanadium in a sample of ore is converted to VO 2+ . The VO 2+ ion is subsequ

Question

The vanadium in a sample of ore is converted to VO2+. The VO2+ ion is subsequently titrated with MnO4- in acidic solution to form V(OH)4+ and manganese(ll) ion. The unbalanced titration reaction is

MnO4-(aq) + VO2+(aq) + H2O(l) ---> V(OH)4+(aq) + Mn2+(aq) + H+(aq)

To tritrate the solution, 59.52 mL of 0.02060 M MnO4- was required. If the mass percent of vanadium in the ore was 59.3%, what was the mass of the ore sample?

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When hydrochloric acid reacts with magnesium metal, hydrogen gas and aqueous magnesium chloride are produced. What volume of 4.9 M HCl is required to react completely with 2.30 g of magnesium?

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A student added 50.0 mL of an NaOH solution to 100.0mL of 0.200 M HCl. The solution was then treated with an excess of aqueous chromium(lll)nitrate, resulting in formation of 2.68 g of precipitate. Determine the concentration of the NaOH solutiion.

Explanation / Answer

multiple questions . please split the rest.

Let the actual mass of vanadium be w.

Thus,

59.52 * 0.02060 * 5 = w/(51)

Thus w = 312.66 mg

Since this is 59.3% of the mass of ore, mass of ore is

= 312.66*100/59.3

= 527.25 mg

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