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Asked by seth

How much limestone (CaCO3) in kilograms would be
required to completely neutralize a lake with a volume of 5.2 x 10^9L . The lake contains 5.0 x 10^-3 moles of H2SO4 per liter?

15 years ago

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Answered by bobpursley
Multiply the volume of the lake by the concentration, that will give you the moles of sulfuric acid. It will take a similar number of moles of limestone.
15 years ago

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