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28g / 55.847g/mol = 0.50 mol

28/18 = 1.5556 (4 decimal places)

27,9 g iron is equal to 0,5 moles.

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Q: How many mol in 28g of iron?

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The answer is 1,25 moles iron(III) oxide.

28g Fe

The mass in grams of 1,40 mol of anhydrous iron(III) sulfate is 559,832.

The reaction is:4 Fe + 3 O2 = 2 Fe2O3The answer is 319,38 g.

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10715600000000000000000

Same as 28g of cast-iron: 1 (One). 454g = 1lb.

The answer is 1,25 moles iron(III) oxide.

Five iron atoms have a mass of 1.3155 x 10-22 grams.

28g Fe

The mass in grams of 1,40 mol of anhydrous iron(III) sulfate is 559,832.

7.2

The reaction is:4 Fe + 3 O2 = 2 Fe2O3The answer is 319,38 g.

Amount of Fe = 55.845/55.845 = 1mol There is 1 mol of Fe in a 55.845g sample. 1 mol of Fe contains 6.02 x 1023 atoms (avogadro constant). Therefore there are 6.02 x 1023 atoms in 55.845g of iron.

5.0x10^25 * (1 mol / 6.022x10^23 atoms) = 83 moles of iron.

The answer is 125,65 g.