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Author Topic: Stoich Chem  (Read 477 times)

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plswhy

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Stoich Chem
« on: May 17, 2018, 08:22:22 PM »

a) If 65 mL of 1.00 M HCl is mixed with 35 mL of 1.00M NaOH, how many moles of each ion are present BEFORE the reaction?

So I understand that HCl + NaOH --> NaCl + H2O and that HCl = 0.065 mol and NaOH= 0.35 mol. I'm just a bit stuck as to how to calculate the individual ions. Please *delete me* Thank you!

Na+

Cl-

H30+

OH-

b) Which reactant is the limiting reagent

c) What are the final concentrations of all the ions present?

d) What is the pH of the final solution.

Thanks so much!
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Borek

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Re: Stoich Chem
« Reply #1 on: May 17, 2018, 08:39:10 PM »

HCl = 0.065 mol and NaOH= 0.35 mol

Check your math, only one of these is OK.

If 0.065 moles of HCl dissociates, how many moles of H+ are produced?
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