How many grams of Fe are needed to create 32.5mol Fe3O4?
1 answer:
Answer:
5444.89 grams of Fe
Explanation:
1) We need the chemical equation for Fe3O4 to find the mole ratio for Fe to Fe3O4. Chemical equation:
3Fe+4H2O → Fe3O4+4H2
The mole ratio of Fe to Fe3O4 is 3:1
2) Now, we find how many grams of Fe are needed to produce 32.5 moles of Fe3O4.
= 97.5 moles Fe needed to produce 32.5 moles of Fe3O4
3) Convert 97.5 moles of Fe into grams. The molar mass of Fe is 55.845g.
= 5444.8875g or 5444.89g.
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Answer:
The heat of the reaction is 105.308 kJ/mol.
Explanation:
Let the heat released during reaction be q.
Heat gained by water: Q
Mass of water ,m= 1kg = 1000 g
Heat capacity of water ,c= 4.184 J/g°C
Change in temperature = ΔT = 26.061°C - 25.000°C=1.061 °C
Q=mcΔT
Heat gained by bomb calorimeter =Q'
Heat capacity of bomb calorimeter ,C= 4.643 J/g°C
Change in temperature = ΔT'= ΔT= 26.061°C - 25.000°C=1.061 °C
Q'=CΔT'=CΔT
Total heat released during reaction is equal to total heat gained by water and bomb calorimeter.
q= -(Q+Q')
q = -mcΔT - CΔT=-ΔT(mc+C)
Moles of propane =
0.0422 moles of propane on reaction with oxygen releases 4.444 kJ of heat.
The heat of the reaction will be: