How many milliliters of 0.100 M HNO3 are needed to neutralize 58.5 mL of 0.0100 M of Al(OH)3 ? Please help I've tried and only have one attempt left.
1 Answer
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Al(OH)3 + 3 HNO3 ----------> Al(NO3)3 + 3 H2O.
0.0585 L * 0.0100 mol/L = 0.000585 moles Al(OH)3.
3 * 0.000585 moles = 0.001755 moles HNO3 reacted.
1000 mL 0.100 M HNO3 contained 0.100 moles.
0.001755 / 0.100 * 1000 mL = 17.6 mL HNO3 needed.