用原子守恒法：
黃鐵礦中硫的質(zhì)量分?jǐn)?shù) = 53.33%
那么1噸礦石中的總硫=1 ×(1-25%)×53.33% = 0.4 t
損失2％之后的硫=0.4 t×(1-2%)=0.392 t
硫酸中硫的質(zhì)量分?jǐn)?shù)為32％
H2SO4質(zhì)量=0.392噸÷32%=1.225 t.
濃硫酸重量質(zhì)量=1.225÷98%=1.25 t怎么求得質(zhì)量分?jǐn)?shù)黃鐵礦FeS2中S的質(zhì)量分?jǐn)?shù) = 32×2 ÷(32×2+56) = 53.33%H2SO4中S的質(zhì)量分?jǐn)?shù) = 32/98 = 32.65 %最后一步有點(diǎn)錯(cuò)誤。改一下：硫酸中硫的質(zhì)量分?jǐn)?shù)為 32.65 %H2SO4質(zhì)量=0.392噸÷32.65%=1.2 t。濃硫酸重量質(zhì)量=1.225÷98%=1.225 t