To find the moles present in 83.20 grams of manganese we must use the molar mass of manganese. This mass is equal to 54.94g/mol. So the moles of manganese (Mn) will be:
[tex]molMn=givengMn\times\frac{1molMn}{MolarMass,gMn}[/tex][tex]molMn=83.20gMn\times\frac{1molMn}{54.94gMn}=1.51molMn[/tex]Answer: 83.20 grams of manganese are equivalent to 1.51 moles
