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How many moles of bicarbonate ions and aluminum ions are present in 65.0 g aluminum bicarbonate?
2 Answers
- Anonymous1 decade agoFavorite Answer
First take the 65.0g of Al(HCO3)3 and convert it to moles of Al(HCO3)3. once you do that just set up a mole to mole conversion.
moles of Al(HCO3)3 X 1 mole Al /1 mole Al(HCO3)3 = mole Al
moles of Al(HCO3)3 X 3 moles HCO3 /1 moles Al(HCO3)3= mole of HCO3
it both conversions the moles of Al(HCO3)3 will cancel in the denominator and you'll be left with the mole of Al or HCO3
Hope this helps!