Question

Un acid anorganic prezinta raportul de masa H:S:O=1:16:32. Se cere formula acidului si cantitatea de acid care reactioneaza cu 200 g azotat de calciu de puritate 82 %

Answer 1

H=1:1=1 =>1:0,5=2
S=16:32=0,5=>0,5:0,5=1
O=32:16=2=>2:0,5=4
=> H2SO4
H2SO4+ Ca(NO3)2 => 2HNO3 + CaSO4
puritatea=mpura*100/mimpura => 82=mpura*100/200=> mpura=164 g
M moleculara H2SO4=98 M Moleculara Ca(NO3)2=40+2*62=40+124=164g
=> M H2SO4=98*164/164=98 g