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Old Apr 16th 2013, 03:27 PM   #1
AileenQ
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Unhappy Help with grams of reactant...

A solution of 150.0 ml of 0.300 M NaOH is mixed with a solution of 200.0 mL of 0.150 M H2SO4.

write a balanced equation and a net ionic reaction.

what is the net ionic product?

how many grams of net ionic product form?

how many grams of the excess reactant is left (in the solution)



I would appreciate any help on how to do this, I have tried already but want to see if I'm going about it correctly.

a. Write a balanced equation & net ionic reaction.
2NaOH (aq) + H2SO4 (aq)-------->Na2SO4(aq)+SO42-(aq)+2H2O(l)
Net ionic: OH-(aq) + 2H+(aq)------>H2O(l)

b. What is the net ionic product?
Would this be H2O...?

c.How many grams of the net ionic product form?
For this i calculated to find the limiting reactant and from my calculations,
NaOH=0.0225 and H2SO4=0.03, so NaOH was my limiting reactant...so then I did:

150.0 mL NaOH x 1liter/1000 ml x 0.300 mol NaOH/1 liter x 2 mol H2O/2 mol NaOH x 18g H2O/1 mol H2O =0.81g H2O

d. How many grams of the excess reactant is left (in the solution)?
I dont know if I just dont understand what it's asking...so my excess reactant would be H2SO4 so would I just subtract my excess (0.03) and my limiting reactant (0.0225)? Or does it have to do with the H2O?

Last edited by AileenQ; Apr 16th 2013 at 07:58 PM. Reason: Tried to work on it even further
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Old Apr 18th 2013, 11:49 AM   #2
Unknown008
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Everything seems good to me

For the last part, you only need to use your limiting reactant. Could you try it out?
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Old Apr 19th 2013, 05:45 AM   #3
KOSNITA
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For the first reaction i doubt about the balancing of the equation and the presence of SO42- on the right??
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