Question

What mass of a solid hydrate of iron (III) sulfate that is 76.3% by mass iron (III) sulfate is needed to supply 1.00 g of iron (III) ion? How many moles of carbon atoms are present in 2.41 x 1022 molecules of acetic acid? How many chloride ions are supplied by 1.00 g of chromium (III) chloride hexahydrate? What mass in grams of cobalt(II) nitrate hexahydrate supplies 0.200 g of cobalt (II) nitrate? What mass in grams of oxalic acid dihydrate supplies 3.01 x 1021 oxygen atoms?

Answer 1

Mass of iron (III) sulfate = 399.88

Mass of Fe₂(SO₄)₃. 7H₂O = 525.88

Mass percentage of Fe₂(SO₄)₃ = 399.88 x 100 / 525.88 = 76.03%

1.00 g of iron (III) ion = 1 /55.85 = 0.01790 Mole

0.01790 Mole of Fe₂(SO₄)₃. 7H₂O = 0.01790 x 525.88 = 9.41 gm

9.41 gm of Fe₂(SO₄)₃. 7H₂O will give you 1 gm of Iron (III) ion

acetic acid have 2 carbons . Each mole of acetic acid have 2 Moles of carbon. Hence let us calculate the mole

2.41 x 10²² molecules of acetic acid =  2.41 x 10²² / 6.023 x 10²³ = 0.04 Moles

No of Moles of carbon = 0.04 x 2 = 0.08 Moles

Mass of chromium (III) chloride hexahydrate = 266.45

No of Moles in 1.00 g of chromium (III) chloride hexahydrate = 1/ 266.45 = 0.00375 Moles

Each Mole of chromium (III) chloride hexahydrate have 3 moles of Chloride ion

Moles of chloride ion = 0.00375 Moles x 3 = 0.011259 Moles

No of Chloride ions = 0.011259 x 6.023 x 10²³ =6.7813 x 10²¹ chloride ions

Mass of cobalt(II) nitrate hexahydrate = 291.03

Mass of cobalt(II) nitrate = 182.943

0.200 g of cobalt (II) nitrate = 0.2 / 182.943 = 0.00109 Moles

0.00109 Moles of cobalt(II) nitrate hexahydrate = 291.03 x 0.00109 = 0.3181 gm

Hence 0.3181 gm of cobalt(II) nitrate hexahydrate supplies 0.200 g of cobalt (II) nitrate

Molecular Weight of oxalic acid dihydrate: 126.064 g/mol

3.01 x 10²¹ oxygen atoms =    3.01 x 10²¹ / 6.023 x 10²³ = 0.0049975 Moles

Each oxalic atom have 6 oxygen atoms . Hence = 0.0049975 Moles / 6 = 8.329 e-4 Moles of oxalic acid presents

8.329 e^-4 Moles of oxalic acid = 8.329 e-4 x 126.064 = 0.10500 gm