Asked by Christie
Given the following equation:
2CH3CO2H + Ca(OH)2 --> Ca(CH3CO2)2 +H2O
If you start with 30 ml of a 5% solution of acetic acid, how much of a .1M Ca(OH)2 solution would it take to completely neutralize the acid?
I'm not quite sure how to start? Should I convert 30 ml to moles? and then use stoichiometry from the equation to convert it to Ca(OH)2?
2CH3CO2H + Ca(OH)2 --> Ca(CH3CO2)2 +H2O
If you start with 30 ml of a 5% solution of acetic acid, how much of a .1M Ca(OH)2 solution would it take to completely neutralize the acid?
I'm not quite sure how to start? Should I convert 30 ml to moles? and then use stoichiometry from the equation to convert it to Ca(OH)2?
Answers
Answered by
DrBob222
Yes. Convert 5% soln to M then M x L = mols and use stoichiometry from there.
Answered by
Christie
How would I convert 5% to M?
Answered by
DrBob222
Is that 5% w/v; if so then it is 5g/100 mL soln. It may be easier to convert to grams. How many g acetic acid are in the 30 mL? That's 5.0 g x (30/100) = ?
Then mols acetic acid = ?/molar mass.
Then mols acetic acid = ?/molar mass.
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