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How many moles of iron are there in 55.85g of Fe3O4

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Answer:

• Molecular mass of Iron (III) tetraoxide


\dashrightarrow \: { \tt{(56 * 3) + (16 * 4)}} \\ = { \tt{168 + 64}} \\ = { \tt{232\:g}}

[ molar masses: Fe → 56, O → 16 ]


\dashrightarrow \:{ \rm{232 \: g \: = 1 \: mole}} \\ \\ \dashrightarrow \: { \rm{55.85 \: g = ( (55.85)/(232)) \: moles }} \\ \\ \dashrightarrow \: { \boxed{ \tt{ = 0.24 \: moles}}}

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