Asked by Grant
What volume (in mL)of 0.0550 M calcium hydroxide is required to neutralize 38.00 mL of 0.0700 M nitric acid?
Answers
Answered by
DrBob222
2HNO3 + Ca(OH)2 ==>Ca(NO3)2 + 2H2O
Moles HNO3 initially = M x L
Moles Ca(OH)2 required is 2 x moles HNO3 (because o the equation 1 mol Ca(OH)2 = 2 moles HNO3).
Then moles Ca(OH)2 = L x M
You know moles and M, calculate L, then multiply by 1000 to convert to mL.
Moles HNO3 initially = M x L
Moles Ca(OH)2 required is 2 x moles HNO3 (because o the equation 1 mol Ca(OH)2 = 2 moles HNO3).
Then moles Ca(OH)2 = L x M
You know moles and M, calculate L, then multiply by 1000 to convert to mL.
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