1. A sample of 24.08g of iron is burnt in air to form an oxide.
a) What is the formula of the oxide if the oxide contains 70% iron by mass?
b) What is the mass of oxygen required for completing the combustion of the iron sample? Correct to 3sig.fig.
c)The oxide is then reduced back to iron by carbon reduction. Caculate the amount of carbon needed to obtain all iron in the product formed from the reaction stated in part b). Correct to 3sig.fig
2. Consider the oxide of metal X, with the formula of XO. Reaction between 123g of XO and hydrogen gas produces metal X and 9.90g of water. Caculate the realtive atomic mass of X
Again F.4 Chemistry
2012-02-07 5:25 am
回答 (2)
2012-02-07 6:57 am
✔ 最佳答案
1.(a)
Relative atomic masses : Fe = 55.8, O = 16
Mole ratio Fe : O
= 70/55.8 : (100 - 70)/16
= 1.254 : 1.875
≈ 2 : 3
Hence, empirical formula = Fe2O3
(b)
Method 1 :
4Fe + 3O2 → 2Fe2O3
No. of moles of Fe = 24.08/55.8 mol
No. of moles of O2 needed = (24.08/55.8) x (3/4) mol
Mass of O2 needed = (24.08/55.8) x (3/4) x 32 = 10.4 g
Method 2 :
Mass of oxygen = 24.08 x [(100 - 70)/70] = 10.3g
(c)
2Fe2O3 + 3C → 4Fe + 3CO2
No. of moles of Fe = 24.08/55.8 mol
No. of moles of C needed = (24.08/55.8) x (3/4) mol
Mass of C needed = (24.08/55.8) x (3/4) x 12 = 3.88 g
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2.
Relative atomic masses : H = 1, O = 16, X = m
In 123 g of XO :
Mass of O = 9.90 x [16/(16 + 1x2)] = 8.8 g
Mass of X = 123 - 8.8 = 114.2g
In XO, mole ratio X : O
114.2/m : 8.8/16 = 1 : 1
114.2/m = 8.8/16
m = 114.2 x (16/8.8) = 207.6
Relative atomic mass of X = 207.6
參考: 土扁
2012-02-07 5:49 am
mass of iron in oxide = 24.08 * 70% = 16.856g
mass of oxygen = 7.224g
iron oxygen
mass 16.856 7.224
no of moles. 16.856/55.6 7.224/16
= 0.30317 = 0.4515
ratio 1 : 1.5
2 : 3
Therefore Fe2O3
mass of oxygen = 7.224g
iron oxygen
mass 16.856 7.224
no of moles. 16.856/55.6 7.224/16
= 0.30317 = 0.4515
ratio 1 : 1.5
2 : 3
Therefore Fe2O3
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