How many grams `H_(2)SO_(4)` can be obtained from `1320gm PbS` as per reaction sequence ?.
`2PbS+O_(2)rarr2PbO+2SO_(2)`
`3SO_(2)+2HNO_(3)+2H_(2)Orarr3H_(2)SO_(4)+2NO`
`["At mass" Pb = 208. S = 32]`.

Explanation

Moles of PbS `= (1320)/(240) = 5.5 mol`
Moles of `SO_(2) = 5.5 mol = "moles"` of `H_(2)SO_(4)`
Mass of `H_(2)SO_(4) = 5.5 xx 98 = 539gm`
[When amount of only one reactant is given generally other is assumed in excess] .