Question 4.15: Calculating the Quantity of a Substance Produced by Reaction......
Calculating the Quantity of a Substance Produced by Reactions Occurring Consecutively
Titanium dioxide, TiO_{2}, is the most widely used white pigment for paints, having displaced most lead-based pigments, which are environmental hazards. Before it can be used, however, naturally occurring TiO_{2} must be freed of colored impurities. One process for doing this converts impure TiO_{2}(s) to TiCl_{4}(g), which is then converted back to pure TiO_{2}(s). The process is based on the following reactions, the first of which generates TiCl_{4}.
2 TiO_{2} (impure) + 3 C(s) + 4 Cl_{2}(g) → 2 TiCl_{4}(g) + CO_{2}(g) + 2 CO(g)
TiCl_{4}(g) + O_{2}(g) → TiO_{2}(s) + 2 Cl_{2}(g)
What mass of carbon is consumed in producing 1.00 kg of pure TiO_{2}(s) in this process?
Analyze
In this calculation, we begin with the product, TiO_{2}, and work backward to one of the reactants, C. The following conversions are required.
kg TiO_{2} → g TiO_{2} → mol TiO_{2} \xrightarrow[]{(a)} mol TiCl_{4} \xrightarrow[]{(b)} mol C → g C
In the conversion from mol TiO_{2} to mol TiCl_{4}, labeled (a), we focus on the second reaction. In the conversion from mol TiCl_{4} to mol C, labeled (b), we focus on the first reaction.
Solve
Using a stepwise approach, we proceed as follows.
Convert from kg TiO_{2} to g TiO_{2} and then to mol TiO_{2} by using the molar mass of TiO_{2}. ? mol TiO_{2} = 1.00 kg TiO_{2} × \frac{1000 g TiO_{2}}{1 kg TiO_{2}} × \frac{1 mol TiO_{2}}{79.88 g TiO_{2}} = 12.5 mol TiO_{2}
Convert from mol TiO_{2} to mol TiCl_{4} by using the stoichiometric factor from the second reaction. ? mol TiCl_{4} = 12.5 mol TiO_{2} × \underbrace{\frac{1 mol TiCl_{4}}{1 mol TiO_{2}}}_{(a)} = 12.5 mol TiCl_{4}
Convert from mol TiCl_{4} to mol C by using the stoichiometric factor from the first reaction. ? mol C = 12.5 mol TiCl_{4} × \underbrace{\frac{3 mol C}{2 mol TiCl_{4}}}_{(b)}= 18.8 mol C
Convert from mol C to g C by using the molar mass of C. ? g C = 18.8 mol C × \frac{12.01 g C}{1 mol C} = 226 g C
The conversions given above can be combined into a single line, as shown below.
? g C = 1.00 kg TiO_{2} × \frac{1000 g TiO_{2}}{1 kg TiO_{2}} × \frac{1 mol TiO_{2}}{79.88 g TiO_{2}} × \overset{(a)}{\frac{1 mol TiCl_{4}}{1 mol TiO_{2}}} × \overset{(b)}{\frac{3 mol C}{2 mol TiCl_{4}}} × \frac{12.01 g C}{1 mol C} = 226 g C
Assess
To obtain the stoichiometric factor for converting from mol TiO_{2} to mol C, we could have reasoned as follows.
From the equations for the two reactions, we see that 3 mol C will give 2 mol TiCl_{4}, which then reacts to give an equal amount (2 mol) of TiO_{2}. That is, 3 mol C will give 2 mol TiO_{2}, or equivalently, 2 mol TiO_{2} requires 3 mol C. Thus, the required stoichiometric factor is (3 mol C)/(2 mol TiO_{2}), the product of the two factors above labeled (a) and (b).