2,5% impuritati => p=97,5%
p/100=mpura/mimpura => mpura=p*mimpura/100
mpura=97,5*600/100=585g NaCl
c/100=md/ms => md=c*ms/100
md=10*5000/100=500g H2SO4
presupunem ca avem cantitatea de sare suficienta
vom determina daca masa de acid este in exces
MNaCl=23+35,5=58,5g/mol
MH2SO4=2+32+64=98g/mol
2*58,5g....98g
2NaCl + H2SO4 -> Na2SO4 + 2HCl
585g........x=?
x=585*98/2*58,5=490g H2SO4
=> 10g H2SO4 in exces
