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Author Topic: Stoich Chem  (Read 578 times)

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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!





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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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?
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation,,, PZWT_s1

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