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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Molarity of solution is mathematically expressed as,
M =
We know that volume = mass/density
Given: mass of solution = 100 g, Density = 1.34 g/ml
∴ volume = 100/1.34 = 88.49 ml = 0.08849 l
Also, we know that molecular weight of sucrose = 342.3 g/mol
∴M =
= 6.979 M
Thus, molarity of solution is 6.979 M
M =
We know that volume = mass/density
Given: mass of solution = 100 g, Density = 1.34 g/ml
∴ volume = 100/1.34 = 88.49 ml = 0.08849 l
Also, we know that molecular weight of sucrose = 342.3 g/mol
∴M =
= 6.979 M
Thus, molarity of solution is 6.979 M