How many grams of ammonium sulfate can be produced if 60.0 mol of sulfuric acid react with an excess of ammonia?

mole ratio of sulfuric acid : ammonium sulphate =  1 :  1
    ∴ if mole of  "          "   = 60.0 mol
     then mole of ammomium sulphate = 60.0 mol

Mass of ammonium sulphate  =  molar mass  *  moles
                                                =   [(2 *1) + (1 * 32) + (4 * 16)]  *  60.0 mol
                                                =  98 g/mol  *  60.0 mol
                                                =  5880 g

Perimeter? wheres the answers