How many mL of 2.0 M KOH would be required to react with (dissolve) 2.38 g of pure aluminum?

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How many mL of 2.0 M KOH would be required to react with (dissolve) 2.38 g of pure aluminum?

2007-07-06 18:11:18 · 1 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry

1 answers

2.38g of Al correspond to 2.38/27=0.081 of Al As Al has a valence of 3 this correspond to 0.081*3=0.243 EQ
KOH is monovalent so you need 0.243mole of KOH
corresponding to 0.243/2=0.1215L on KOH 2M

121.5mL is the answer

2007-07-06 18:36:15 · answer #1 · answered by maussy 7 · 0⤊ 0⤋