含FeS2 80%的硫铁矿200t,灼烧时硫损失2%,能制98.3%的硫酸多少吨

问题描述:

含FeS2 80%的硫铁矿200t,灼烧时硫损失2%,能制98.3%的硫酸多少吨

200*80%*64/120=85.333吨硫损失2%=85.333*2%=1.706785.333-1.70667=83.626S+O2=SO22SO2+O2=2SO3H2O+SO3=H2SO42S+3O2+2H2O=H2SO464 98 83.626 X=128.053128.053/98.3%=130...