Answer:
• Molecular mass of Iron (III) tetraoxide
[tex] \dashrightarrow \: { \tt{(56 \times 3) + (16 \times 4)}} \\ = { \tt{168 + 64}} \\ = { \tt{232\:g}}[/tex]
[ molar masses: Fe → 56, O → 16 ]
[tex] \dashrightarrow \:{ \rm{232 \: g \: = 1 \: mole}} \\ \\ \dashrightarrow \: { \rm{55.85 \: g = ( \frac{55.85}{232}) \: moles }} \\ \\ \dashrightarrow \: { \boxed{ \tt{ = 0.24 \: moles}}}[/tex]
