F4 CHEM (reacting masses)

2011-02-19 8:36 am

1. Hydrated iron(II) sulphate was heated in and the liquid was collected in another test tube through a delivery tube.
a. Explain why the tube containing FeSO4 was clamped in a slanting positioin.
b. Suggest a test to show that the liquid collected was water.
c. Why is it desirable to keep the free end of the delivery tube above the liquid collected in the test tube?
d. The formula for hydrated iron(II) sulphate is FeSO4 xH2O. In the above experiment, 30.6 g of the sulphate produced 13.9 g of water. Calculate the value of x.
(Rel. atm. masses: H = 1.0, O = 16.0, S = 32.1, Fe = 55.8)

2a. When iron(III) hydroxide was heated to a high temperature, iron(III) oxide was formed.
2Fe(OH)3 (s) -----> Fe2O3 (s) + 3H2O (g)
Calculate the maximum mass of water formed from 5.35 g of iron(III) hydroxide.

2b. Another oxide of iron consists of 72.4% iron by mass. Calculate the empirical formula of this oxide.
(rel. atm. masses: H = 1.0, O = 16.0, Fe = 55.8)

3. Aluminium can displace copper from copper(II) sulphate solution.
a. Write a chemical equation for the reaction involved.
b. In a certain experiment, 2.58 g of copper were produced from 1.61 g of aluminium and excess copper(II) sulphate solution. What was the percentage yield of copper in the reaction?
(rel atm masses: Al = 27.0, Cu = 63.5)

4. Manganese(IV) oxide reacts with concentrated hydrochloric acid according to the following equation:
MnO2 (s) + 4HCl (aq) ----------> MnCl2 (aq) + Cl2 (g) + 2H2O (l)
217 g of manganese(IV) oxide were treated with hydrochloric acid containing 274 of hydrogen chloride. Determine:
a. the limiting reactant
b. the mass of chlorine produced
c. the mass of excess reagent left after the reaction
(rel atm masses: H = 1.0, O = 16.0, Cl = 35.5, Mn = 54.9)

回答 (1)

pingshek

2011-02-19 9:32 am

✔ 最佳答案

1. a.
This is to prevent the condensed liquid water from going back to the test tube. The cool water may crack the hot tube.

1. b.
Water changes dry cobalt(II) chloride from blue to pink.

1. c.
This is to avoid sucking back.

1. d.
Mole ratio FeSO4 : H2O = 1 : x
(30.6 - 13.9)/(55.8 + 32.1 + 16x4) : 13.9/(1x2 + 16) = 1 : x
16.7/151.9 : 13.9/18 = 1 : x
(16.7/151.9)x = 13.9/18
x = 7

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2. a.
2Fe(OH)3(s) → Fe2O3(s) + 3H2O(g)
No. of moles of Fe(OH)3 = 5.35/(55.8 + 16x3 + 1x3) = 0.0501 mol
Max. no. of moles of H2O formed = 0.0501 x (3/2) = 0.0752 mol
Max. mass of H2O formed = 0.0752 x (1x2 + 16) = 1.35 g

2. b.
Mole ratio Fe : O
= 72.4/55.8 : (100 - 72.4)/16
= 1.3 : 1.725
= 3 : 4
Hence, empirical formula = Fe3O4

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3. a.
2Al + 3CuSO4 → Al2(SO4)3 + 3Cu

3. b.
No. of moles of Al used = 1.61/27 = 0.0596 mol
Max. no. of moles of Cu formed = 0.0596 x (3/2) = 0.0894 mol
Max. mass of Cu formed = 0.0894 x 63.5 = 5.68 g
% yield of Cu = (2.58/5.68) x 100% = 45.4%

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4.
(a)
MnO2 + 4HCl → MnCl2 + Cl2 + 2H2O
No. of moles of MnO2 added = 217/(54.9 + 16x2) = 2.50 mol
No. of moles of HCl = 274/(1 + 35.5) = 7.51 mol

For complete reaction of 2.50 mol of MnO2,
no. of moles of HCl needed = 2.50 x 4 = 10 mol
Hence, HCl is the limiting reactant.

(b)
No. of moles of HCl reacted = 7.51 mol
No. of moles of Cl2 produced = 7.51 x (1/4) = 1.88 mol
Mass of Cl2 produced = 1.88 x (35.5x2) = 133 g

參考: miraco

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