Question

83.20 grams of manganese to moles

Answer 1

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:

`molMn=givengMn*(1molMn)/(MolarMass,gMn)`
`molMn=83.20gMn*(1molMn)/(54.94gMn)=1.51molMn`

Answer: 83.20 grams of manganese are equivalent to 1.51 moles