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mojhsa [17]
2 years ago
14

How many fe(ii) ions are there in 10.0 g of feso4?

Chemistry
2 answers:
Eva8 [605]2 years ago
6 0

The number of Fe(ii) ions that are in 10.0g of FeSO4  is 3.961 x10^22 ions

calculation

find the number of moles of FeSO4 = mass/molar mass

= 10g / 151.9 g/mol=0.0658 moles  since there 1 atom of fe the moles of Fe= 0.0658 moles

by use of Avogadro's constant

that is  1 mole= 6.02 x10^23 ions

          0.0658moles=? ions


= 0.0658 mole x(6.02 x10^23)] / 1 mole = 3.961 x10^22 ions

hram777 [196]2 years ago
3 0

Iron sulfate can be shown as FeSO4 and the number mole of FeSO_4  can be determined as

Molar mass of Iron sulfate = 151 g /mole

Mass of Iron sulfate = 10 g

Mole = mass /molar mass\\ = 10 /151 = 0.066 mol

Number of Fe ion  can be determined as

Mole = Number of ion / Avogadro number\\ 0.066 = Number of ion / 6.02 * 10^23\\ Number of ion = 3.98 * 10^22

Thus, 3.98 * 10^22 ions are present in Iron sulfate

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