(a) Describe briefly a suitable procedure for preparing a pure sample of MgSO\(_4\) starting from MgO.
(b)(i) Mention two sources of water pollution.
(ii) Explain why the sample of air collected in the process of boiling water is richer in oxygen than atmospheric air
(iii) Mention one substance used as coagulant in water treatment plants.
(c)(i) State two physical porperties of chlorine.
(ii) Write an equation to show how chlorine reacts with iron
(iii) Why is Chlorine preferred to sulphur (IV) oxide in the bleaching of cotton
(d) Bleaching powder reacts with dilute HCl according to the reaction below;
CaOCl\(_{2(s)}\) + 2HCI\(_{(aq)}\) -> CaCl\(_{2(aq)}\) + H\(_2\)O\(_{(l)}\) + Cl\(_{2(g)}\)
Calculate the mass of bleaching powder that will produce 400cm\(^3\) of chlorine at 25\(^o\)C and a pressure of 1.20 x 10\(^5\) NM\(^{-2}\). [O = 16.0; Cl = 35.5; Ca = 40.0;1 mole of gas occupies 22.4 dm\(^3\) at s.t.p; standard pressure = 1.01 x 10\(^6\) Nm\(^{-2}\)]
(a) MgSO\(_4\) is prepared in the laboratory by the action of dilute H\(_2\)SO\(_4\) on MgO. On crystallisation the hepta hydrate crystals MgSO\(_4\). 7H\(_2\)O known as Epsom salt is obtained. MgO + H\(_2\)SO\(_4\) -> MgSO\(_4\) + H\(_2\)O.
(b)(i) - Industrial effuents -.Human waste. -Human activities
(ii) During boiling some molecules of water do break up into oxygen and hydrogen, the oxygen then combine with the atmospheric oxygen.
(iii) Alum (a double salt)
(c)(i) - It is a greenish yellow gas
- It is moderately soluble in water
- It is about 2.5 times denser than air
-It can be liquefied under pressure
- It is poisonous.
(ii) 2 FeCl\(_2\) + Cl\(_2\) ----> 2FeCI\(_3\)
(iii) Chlorine is preferred to sulphur (IV) oxide in the bleaching of cotton because the bleaching action of SO\(_2\) usually does not last since the bleached dye become re-oxidized by atmospheric oxygen to form the original coloured compound. This is not so with chlorine.
(d) CaOCl\(_2\) + 2HCI -> CaCl\(_2\) + H\(_2\)O + Cl\(_2\) from the equation.
(40 + 16 + 35.5 x 2)g CaOCl\(_2\) will produce 22400cm\(^3\) of Cl\(_2\) at s.t.p.
127g -> 22400cm\(^3\) at temp.
= 273k and pressure
= 1.01 x 10\(^5\) N/M\(^{-2}\); Volume of CaOCl\(_2\) at s.t.p.
\(\frac{P_1 V_1}{T_1}\) = \(\frac{P_2V_2}{T_2}\)
\(V_2 = \frac{P_1V_1T_2}{T_1P_2}\)
= \(\frac{1.20 \times 10^5 \times 400 \times 273}{298 \times 1.01 \times 1.05}\)
= 435.4cm\(^3\)
127g -> 22400dm\(^3\) ;
y -> 435.4dm\(^3\)
y mass or CaOCl\(_2\) = \(\frac{435.4}{22400}\) x 127 = 2.468g
Contributions ({{ comment_count }})
Please wait...
Modal title
Report
Block User
{{ feedback_modal_data.title }}