Asked by Kim
What is the minimum volume of 2.00 M hydrochloric acid necessary to neutralize completely the hydroxide in 800.0 mL of 0.3 M calcium hydroxide?
Answers
Answered by
Damon
2HCl + Ca(OH)2 ---> 2 H2O
.8 liters *.3 mol/liter = .24 mol Ca(OH)2
so
we need .48 mols HCl
.48 mol * 1000 ml/2 mol = 240 mL
.8 liters *.3 mol/liter = .24 mol Ca(OH)2
so
we need .48 mols HCl
.48 mol * 1000 ml/2 mol = 240 mL
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