Respuesta :
[tex]2NH_{3} + H_{2} SO _{4} ----\ \textgreater \ (NH _{4} )_{2} SO _{4} [/tex]
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
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
How many grams of ammonium sulfate can be produced if 60.0 mol of sulfuric acid react with an excess of ammonia? D. 7,930 g