BaCl2+H2SO4==BaSO4+2HCl
n(BaCl2)=0.05*0.02=0.001mol
n(H2SO4)=15*0.1/98=0.015mol
所以H2SO4过量,以B aCl2计算,则n(BaSO4)=n(BaCl2)=0.001mol
所以,m(BaSO4)=0.001molx233=0.233g
V(H2SO4)=15/1.07=14ml
剩余n(H2SO4)=0.015-0.001=0.014mol
故c(H2SO4)=0.014/(50+14)*1000=0.22mol/L