Thank you Calculate the molar mass of ALUM, KAl(So_4)_2 middotI2H_2O or potassiu

Question

Thank you Calculate the molar mass of ALUM, KAl(So_4)_2 middotI2H_2O or potassium aluminum sulfate dodecahydrate. Be sure to include the twelve waters of hydration in the molar mass Use the overall reaction and the fact that we use 12 mL of 4 M KOH and 15 mL of 6 M H_2SO_4 for this experiment to prove that aluminum metal is the limiting reagent in this experiment To do this calculate the amount of aluminum metal that could be reacted by using the steps below: Calculate the moles of KOH and H_2SO_4 added to the reaction mixture from the data above. Calculate the number of moles of A1 that would be required to react the above moles of the two reactants. Calculate the minimum mass of aluminum that could be reacted by the reagents above. How does the number you calculated in part (c) prove that aluminum is the limiting reagent for this experiment?

Explanation / Answer

Moles of KOH = Volume of KOH in L * Molarity 0.012L * 4M = 0.048 mol

Moles of H2SO4 = Volume of H2SO4 in L * Molarity = 0.015 L *6M =0.09 mol

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Overall reaction : K+ +   Al ^3+   + 2SO4^2- + 12H2O -----> KAl(SO4)2 .12H2O

1 mol Al will react with 1 mol of KOH and 2 mol of H2SO4.

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Al is present in the solution in smallest amount and it will be completely consumed. So, this is the limiting reagent

Minimum moles of Al = 0.048

mass of Al = 1mol * 26.98 g/mol = 26.98 g

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