A 0.2412 gram sample of aluminum ore was dissolved in a dilute mineral acid. The solution was treated with an excess of sodium oxalate. The resultant aluminum oxalate precipitate was filtered, washed and redissolved in mineral acid, then titrated with 42.12 mL of 0.1098 N KMnO4. Calculate the percent of aluminum in the sample.
4Al + 6Na2(C2O4) ---- 2Al2(C2O4)3 + 6KMnO4 ---- 3K2(C2O4) Aluminum oxalate and potassium permanganate are at a 1:3 ratio.... Moles of KMnO4 C = 0.1098 mol/L v = 42.12 ml = 0.04212 L C = n/v n= CV = 0.004624776 mols Therefore moles of aluminum oxylate = 0.001541592 mol = moles of aluminum m= 26.98 g/mol * moles mass of aluminum in oxylate = 0.04159215 g % of initial sample = mass of aluminum / mass of ore = 17.24% aluminum in ore sample Perhaps check my math but the method is for sure correct.
determine which you do catch the Hydrogen gas in a balloon. combination Lye with water in a pitcher bottle (Coke liter bottle is huge). drop in Aluminum foil (do no longer give way) and cap off with a balloon. The balloon will fill with hydrogen. do away with the balloon, pinching the backside keeping the hydrogen; shop including Aluminum foil strips and place the balloon lower back over the precise till at last the balloon is of a sturdy length. Tie the balloon off, tape a fuse (some thing that burns slow) to the balloon, permit it bypass up interior the air. whilst the hearth its the balloon, it pops and the hydrogen + oxygen (interior the air) + spark reasons a small flash (hydrogen explosion). Do it at night for a extra advantageous view.
