If a student weighs out 0.482 grams of aluminum foil and dissolves it in KOH solution, how many moles of aluminum metal react?
How many moles of KOH are necessary to react with the aluminum metal?
Calculate the number of moles of KOH in 13.0 mL of 3.0 M KOH solution?
2Al(s) + KOH + 6H2O -------------> 2KAl(OH)4 + 3H2
no of moles of Al = W/G.A.Wt
= 0.482/27 = 0.0178moles
no of moles of KOH = molarity * volume in L
= 3*0.013 = 0.039 moles
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