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Year 12 Chemistry Test 1 Solutions
Mercy College
YEAR 12 CHEMISTRY TEST 1 SOLUTIONS
1c, 2c, 3b, 4d, 5d, 6a, 7b, 8d, 9a, 10d
11. Calculated the density of a saturated solution of 10.7% K2Cr2O7 by mass which has a
concentration of 0.39 molL-1 at 20oC.
[4 marks]
Let’s take 1 L of the solution. Hence,
V=1L
c(K2Cr2O7 ) = 0.39 molL-1
n(K2Cr2O7 ) = c x V = 0.39 x 1 = 0.39 mol
(1 mark)
m(K2Cr2O7 ) = n(K2Cr2O7 ) x M(K2Cr2O7 )
= 0.39 mol x 294.1846
= 114.73g
(1 mark)
%(K2Cr2O7 ) = [m(K2Cr2O7 )/m(solution)] x 100 L
10.7 = [114.73 / m(solution)] x 100
m(solution) = 1072.26g
(1 mark)
Density(solution) = (mass of solution)/(volume of solution)
= 1072.26/1
= 1072.26 g/L
(1 mark)
The density of the saturated solution is 1.07 g/ml
12. A particular ore contains 85% PbCO3(s) by mass. In the pyrometallurgical industry, PbO is
the desired solid from the ore. To produce this, the ore is heated so that the PbCO3 reacts
as follows:
PbCO3(s)
PbO(s) + CO2(g)
The PbO is further treated to produce pure lead metal.
What mass of ore is needed to form 1000 kg of the lead metal?
[4 marks]
n( Pb ) = m( Pb )/M( Pb ) = 1.00 x 106 /207.20
= 4826.25
(1 mark)
n( PbCO3 ) = n( Pb ) = 4826.25
(1 mark)
m( PbCO3 ) = n( PbCO3 ) x M( PbCO3 )
= 4826.25 x 267.21
= 1.2896 x 106 g
(1 mark)
%( PbCO3 in ore) = [m( PbCO3 ) / m(ore)] x 100
85
= [1.2896 x 106 / m(ore)] x 100
Therefore m(ore) = 1.517 x 106 g
m(ore) = 1517 kg
24th February 2004
(1 mark)
Page 1 of 3
Cesareo
Year 12 Chemistry Test 1 Solutions
Mercy College
13. A 40 litre cylinder at 7oC contains ammonia at 170 kPa and oxygen at 62.4 kPa. The
system is ignited and the following reaction occurs:
4NH3(g) + 3O2(g)
6H2O(g) + 2N2(g) .
After the reaction the gases are pumped into a moisture free cooling vessel. The vessel
cools its contents to the point where all the gases turn into their solid phase. Calculate the
mass of the final solid mixture.
[12 marks]
n(NH3(g) ) = (PV)/(RT)
= (170 x 40)/(8.315 x 280)
= 2.9207
(1 mark)
n(O2(g) ) = (PV)/(RT)
= (62.4 x 40)/(8.315 x 280)
= 1.0721
(1 mark)
Determination of Limiting Reagent (LR)
Choose one the reactants. I choose NH3(g)
n(NH3(g) ) = 2.9207
The theoretical (TH) number of moles of O2(g) needed so that the NH3(g) is
completely consumed = ¾ n(NH3(g) ) [from equation]
= ¾ x 2.9207
= 2.1905
(1 mark)
Hence, n(O2(g) )TH = 2.1905
Compare the theoretical amount with the real amount:
n(O2(g) )TH = 2.1905 & n(O2(g) ) = 1.0721
(1 mark)
n(O2(g) )TH > n(O2(g) ). Hence, there is not enough O2(g) to react with all the
(1 mark)
NH3(g) . Therefore O2(g) is the LR
n(H2O(g) ) = 2n(O2(g) ) = 2 x 1.0721 = 2.1441
n( N2(g) ) = (2/3) n(O2(g) ) = (2/3) x 1.0721 = 0.7147
n(NH3(g) )unreacted = n(NH3(g) ) - n(NH3(g) )reacted
= 2.9207 – [ (4/3)n( O2(g) )
= 2.9207 – [ (4/3) x 1.0721]
= 1.4913
n(H2O(s) ) = n(H2O(g) )
= 2.1441
n( N2(s) ) = n( N2(g) )
= 0.7147
n( NH3(s) ) = n(NH3(g) )unreacted = 1.4913
m( H2O(s) ) = n( H2O(s) ) x M( H2O ) = 2.1441 x 18.0152
= 38.627 g
m( N2(s) ) = n( N2(s) ) x M( N2 )
(1 mark)
(1 mark)
(1 mark)
(1 mark)
= 0.7147 x 28.0143
= 20.022 g
(1 mark)
m( NH3(s) ) = n( NH3(s) ) x M( NH3 ) = 1.4913 x 17.0305
= 25.397 g
(1 mark)
Mass of the final solid mixture = m( H2O(s) ) + m( N2(s) ) + m( NH3(s) ) = 84.0 g
(1 mark)
24th February 2004
Page 2 of 3
Cesareo
Year 12 Chemistry Test 1 Solutions
Mercy College
14. Jeremy wants to determine the percentage by mass of sodium, carbon and oxygen in a
compound that has only these three types of atoms. He takes a sample of the dry compound
but before he gets to the balance he accidentally spills a small volume of pure water on the
sample. He knows that the compound does not react with water so he decides not to dry it.
Instead he commences his analytical procedure for determining the sodium, carbon and
oxygen atoms in the compound. He therefore adds excess barium chloride solution to 1.502
g of the wet compound. The solid barium carbonate residue that is formed is filtered out
and is found to have a mass of 1.545 g. Jeremy then takes the sodium chloride filtrate
solution and adds 31.31 ml of 0.500 mol L-1 silver nitrate solution which is just the right
volume that causes all the sodium chloride to form a residue of solid silver chloride.
Calculate for Jeremy the percentage by mass of sodium, carbon and oxygen in the DRY
compound.
[18 marks]
n(BaCO3(s) ) = m(BaCO3(s) )/M(BaCO3(s) ) = 1.545/197.3 = 7.831 x 10-3
(1 mark)
n(O)in dry compound = 3 x n(BaCO3(s) ) = 3 x 7.831 x 10-3 = 2.349 x 10-2
(1 mark)
-2
-1
m(O)in dry compound = n(O) x M(O) = 2.349 x 10 x 16.00 = 3.759 x 10 g
(1 mark)
%(O)in wet compound = [m(O)/m(wet compound)] x 100 = [(3.759 x 10-1)/ 1.502] x 100 = 25.0
(1 mark)
(1 mark)
(1 mark)
-3
n(C)in dry compound = n(BaCO3(s) ) = 7.831 x 10
m(C)in dry compound = n(C) x M(C) = 7.831 x 10-3 x 12.01 = 9.405 x 10-2 g
%(C)in wet compound = [m(C)/m(wet compound)] x 100 = [(9.405 x 10-2) / 1.502] x 100 = 6.26
(1 mark)
(1 mark)
n(AgNO3(aq) ) = c(AgNO3(aq) ) x V(AgNO3(aq) ) = 0.500 x 0.03131 = 1.565 x 10-2
NaCl(aq) + AgNO3(aq)
AgCl(s) + NaNO3(aq)
n(NaCl(aq) ) = n(AgNO3(aq) ) = 1.565 x 10-2
n(Na)in dry compound = n(NaCl(aq) ) = 1.565 x 10-2
(1 mark)
-2
-1
(1 mark)
(1 mark)
m(Na)in dry compound = n(Na) x M(Na) = 1.565 x 10 x 22.99 = 3.600 x 10 g
%(Na)in wet compound = [m(Na)/m(wet compound)] x 100 = [(3.600 x 10-1) / 1.502] x 100
= 24.0
(1 mark)
%(H2O ) = 100 – 25.0 – 6.26 – 24.0
= 44.74
(1 mark)
m(H2O ) = (44.74/100) x 1.502 = 0.6720 g
m(dry compound) = 1.502 – 0.6720 = 0.830 g
(1 mark)
(1 mark)
%(O)in dry compound = [ (m(O)in dry compound) / 0.830 ] x 100 = [(3.759 x 10-1) / 0.830] x 100
= 45.3
(1 mark)
%(C)in dry compound = [ (m(C)in dry compound) / 0.830 ] x 100 = [(9.405 x 10-2) / 0.830] x 100
= 11.3
%(Na)in dry compound = 100 – 45.3 – 11.3 = 43.4
24th February 2004
(1 mark)
(1 mark)
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