|
Calculations
on the mole concept
Francine
Taylor-Campbell, Contributor
 |
| Portrait
of a rural slum. - Ian Allen/Staff Photographer | Good
day. Let us try this question. 1.
(a) On complete combustion, 4.0g of a hydrocarbon give 11.0 g of carbon dioxide
and 9.0 g of water. What is the formula of the hydrocarbon? b)
1.89g of zinc nitrate was heated according to the equation 2Zn(NO3)2
(s) == 2ZnO (s) + 4NO2 (g) + O2 (g)
What
is the percentage by mass of oxygen in zinc nitrate? What volume of oxygen at
r.t.p. is given off when 1.89g of zinc nitrate are heated? ANSWERS
(a)
M r of CO2 = 12+(16*2) = 44
M
r of H2O = (1*2)+16 = 18
#mol
CO2 = 11 /44 = 0.25 mol
#mol
H2O = 9/18 = 0.5 mol
Based
on the formula of the hydrocarbon CxHy
#mol
of C = 0.25 mol and #mol of H = 2*0.5 = 1 mol
Ratio
of C:H = 0.25: 1 = 1:4
Hydrocarbon
is CH4 (b)
2Zn(NO3)2 == 2ZnO + 4NO2
+ O2 Molar
mass of Zn(NO3)2 = 65 + 14*2 + 16*6 = 189g % O = mass
of oxygen in zinc nitrate = (96/189)*100 = Total mass of zinc nitrate 50.8%
(ii)
1.89g of Zn(NO3)2 represents 1.89/189
= 0.01 mol According
to the equation, 2 mol of Zn(NO3)2
gives off 1 mol O2 Therefore
0.01 mol Zn(NO3)2 will produce 0.01/2
= 0.005 mol O2 At
r.t.p. 1 mole of any gas occupies a volume of 24dm 3 0.005
mol O2 has a volume of 0.005*24 = 0.12dm 3 Question
2 2.
Ammonia and carbon dioxide react to form water and a solid, urea, CON2H4. In the
reaction, 72dm³
of carbon dioxide at r.t.p. are converted to urea. Write
the equation for the formation of urea. Calculate
the volume of ammonia at r.t.p. which reacted. Calculate the mass of urea formed.
Answers:
(i
) 2NH3 (g) + CO2 (g) == H2O
(l) + CON2H4 (s) (ii)
Based on the equation 2 mol NH3 react with 1 mol CO2 (2:1
ratio) Hence
if 72 dm³
of CO2 is used then 2 x 72 dm³
of NH3 would react. Volume
of ammonia = 144dm³ (iii)
1 mol of gas at r.t.p has a volume of 24dm³
#
mol CO2 used = 72/24 = 3 mol Using
the equation again 1 mol CO2 produces 1 mol CON2H4 (urea)
Thus
3 mol CO2 produces 3 mol urea Molar
mass of urea = (12 + 16 + 14*2 + 1*4) = 60g 3
mol urea has a mass of 60 x 3 = 180g. Now,
please attempt this question. 3.
Give the equation for the reaction between methane and steam. Calculate
the maximum volume of hydrogen, measured at s.t.p., which can be obtained from
16g of methane. Answers
CH4
(g) + H2O (g) — CO (g) + 3H2
(g) M
r of CH4 = 12+ (1*4) = 16 #mol
CH4 = 16/16 = 1 mol Based
on reaction 1 mol of CH4 produces 3 mol of H2
gas At
s.t.p 1 mol of any gas = 22.4 dm-3
Volume
of 3 mol H 2 = 3*22.4 = 67.2 dm-3
Francine
Taylor-Campbell is an independent contributor. | 
