Help please!.So 700 ml 84.5 % H2SO4 (p=1.78 g/cm3) react with 1200 ml 42% NaOH (p=1.45 g/cm3).How much ml needed 35% KOH (p=1.35) to neutralized getted acid from previous reaction! em sorry for my bad english@

Question

vincent-lyon.fr · Answer

You have two things to do before finding the amount of KOH needed.

1. Write equation of reaction between sulfuric acid and sodium hydroxide.
2. Find amounts reacting, and how much acid is in excess.

anonymous · Answer

okey equation is H2SO4+2NAOH=NA2SO4+2H20

vincent-lyon.fr · Answer

Great!

anonymous · Answer

amount nH2SO4=10.73mol
amount nNAOH=18.26 mol

anonymous · Answer

but i dont know is it right

vincent-lyon.fr · Answer

Go on. If I do not answer right away, it is because I am working on your question ;-)

vincent-lyon.fr · Answer

I find 8.75 mol H2SO4
Let's work out that together.
How did you do it?

anonymous · Answer

so

anonymous · Answer

firstly i find mollar concentration

anonymous · Answer

right? cuz i have p

vincent-lyon.fr · Answer

I agree with your amount of NaOH :)

anonymous · Answer

and volume

anonymous · Answer

hm i cant find mistake ((

vincent-lyon.fr · Answer

I think you were right. Wait a second, I'll confirm.

vincent-lyon.fr · Answer

You are absolutely right!

Now, find what is left after the reaction is over.

anonymous · Answer

emm there i have problem i dont know how :D

vincent-lyon.fr · Answer

Are you here?

anonymous · Answer

yes

vincent-lyon.fr · Answer

Follow the link I sent you in the private message.

anonymous · Answer

ye i join