neutralne soli

look's picture

Koliko molova neutralne soli moze da se dobije djelovanjem natrijum hidroksida u visku, na 250 mL 0.1 mol/L rastvora sumporne kiseline?

buduci_student's picture

Imas li resenje?

look's picture

0.025

look's picture

0.025

buduci_student's picture

Imam neku ideju,mozda ovako

c=n/V

n=c*V=0,1mol/dm3* 0,25 dm3=0,025 Sumporne kiseline

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2NaOH+H2SO4---------->Na2SO4 +2H2O

1 mol H2SO4 ----> 1 mol Na2SO4

0,025 mol H2SO4----> xmol Na2SO4

x= 0,025 mol Na2SO4

 

buduci_student's picture

WOooooooooooHHHHHHHHooooooooo tacan je, toliko sam se obradovao, ne mozes i sam zamisliti :D