ile gramów wodorotlenku potasu całkowicie zobojetni 10g kwasu siarkowego (VI)
mKOH=56u
mH2SO4=98u
2KOH + H2SO4---->K2SO4 + H2O
2*56g KOH-------98g H2SO4
xg KOH------------10g H2SO4
x = 11,4g KOH
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mKOH=56u
mH2SO4=98u
2KOH + H2SO4---->K2SO4 + H2O
2*56g KOH-------98g H2SO4
xg KOH------------10g H2SO4
x = 11,4g KOH