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Chapter 4: Problem 72

Titanium tetrachloride, \(\mathrm{TiCl}_{4}\) is prepared by the reaction below. $$\begin{aligned} &3 \mathrm{TiO}_{2}(\mathrm{s})+4 \mathrm{C}(\mathrm{s})+6 \mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow 3 \mathrm{TiCl}_{4}(\mathrm{g})+2 \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{CO}(\mathrm{g}) \end{aligned}$$ What is the maximum mass of \(\mathrm{TiCl}_{4}\) that can be obtained from \(35 \mathrm{g} \mathrm{TiO}_{2^{\prime}} 45 \mathrm{g} \mathrm{Cl}_{2^{\prime}}\) and \(11 \mathrm{g} \mathrm{C} ?\)

Short Answer

Expert verified
The maximum mass of \(TiCl4\) that can be obtained is 30 g.

Step by step solution

01

Convert given mass to moles

Molar mass of \(TiO2\) is about 79.9 g/mol. Moles of \(TiO2\) = 35 g / 79.9 g/mol = 0.437 mol. Molar mass of \(Cl2\) is about 71 g/mol. Moles of \(Cl2\) = 45 g / 71 g/mol = 0.633 mol. Molar mass of \(C\) is about 12 g/mol. Moles of \(C\) = 11 g / 12 g/mol = 0.917 mol.
02

Identify the limiting reactant

Looking at the balanced chemical equation, for every 3 moles of \(TiO2\), 6 moles of \(Cl2\) and 4 moles of \(C\) are required. So, normalized for \(TiO2\): \(TiO2\) = 0.437 mol, \(Cl2\) = 0.633 mol / 2 = 0.316 mol, \(C\) = 0.917 mol / 1.33 = 0.69 mol. In conclusion, \(Cl2\) is the limiting reactant because it has the smallest mole quantity.
03

Calculate the maximum mass of \(TiCl4\)

The balanced equation tells us that for every 6 moles of \(Cl2\), 3 moles of \(TiCl4\) can be formed. Therefore, moles of \(TiCl4\) = moles of \(Cl2\) / 2 = 0.316 mol / 2 = 0.158 mol. The molar mass of \(TiCl4\) is 189.7 g/mol. So, the maximum mass of \(TiCl4\) is = moles of \(TiCl4\) * molar mass of \(TiCl4\) = 0.158 mol * 189.7 g/mol = 30 g.

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Most popular questions from this chapter

If \(46.3 \mathrm{g} \mathrm{PCl}_{3}\) is produced by the reaction $$ 6 \mathrm{Cl}_{2}(\mathrm{g})+\mathrm{P}_{4}(\mathrm{s}) \longrightarrow 4 \mathrm{PCl}_{3}(\mathrm{l}) $$ how many grams each of \(\mathrm{Cl}_{2}\) and \(\mathrm{P}_{4}\) are consumed?

The reaction of potassium superoxide, \(\mathrm{KO}_{2}\), is used in life- support systems to replace \(\mathrm{CO}_{2}(\mathrm{g})\) in expired air with \(\mathrm{O}_{2}(\mathrm{g}) .\) The unbalanced chemical equation for the reaction is given below. $$\mathrm{KO}_{2}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g}) \longrightarrow \mathrm{K}_{2} \mathrm{CO}_{3}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g})$$ (a) How many moles of \(\mathrm{O}_{2}(\mathrm{g})\) are produced by the reaction of \(156 \mathrm{g} \mathrm{CO}_{2}(\mathrm{g})\) with excess \(\mathrm{KO}_{2}(\mathrm{s}) ?\) (b) How many grams of \(\mathrm{KO}_{2}(\mathrm{s})\) are consumed per \(100.0 \mathrm{g} \mathrm{CO}_{2}(\mathrm{g})\) removed from expired air? (c) How many \(\mathrm{O}_{2}\) molecules are produced per milligram of \(\mathrm{KO}_{2}\) consumed?

A drop \((0.05 \mathrm{mL})\) of \(12.0 \mathrm{M} \mathrm{HCl}\) is spread over a sheet of thin aluminum foil. Assume that all the acid reacts with, and thus dissolves through, the foil. What will be the area, in \(\mathrm{cm}^{2}\), of the cylindrical hole produced? (Density of \(\mathrm{Al}=2.70 \mathrm{g} / \mathrm{cm}^{3} ;\) foil thickness \(=0.10 \mathrm{mm} .)\) \(2 \mathrm{Al}(\mathrm{s})+6 \mathrm{HCl}(\mathrm{aq}) \longrightarrow 2 \mathrm{AlCl}_{3}(\mathrm{aq})+3 \mathrm{H}_{2}(\mathrm{g})\)

Consider the chemical equation below. What is the maximum number of moles of \(\mathrm{K}_{2} \mathrm{SO}_{4}\) that can be obtained from a reaction mixture containing 5.0 moles each of \(\mathrm{KMnO}_{4}, \mathrm{KI},\) and \(\mathrm{H}_{2} \mathrm{SO}_{4} ?\) (a) \(3.0 \mathrm{mol}\); (b) \(3.8 \mathrm{mol}\) (c) \(5.0 \mathrm{mol} ;\) (d) \(6.0 \mathrm{mol} ;\) (e) \(15 \mathrm{mol}\). $$\begin{array}{r} 2 \mathrm{KMnO}_{4}+10 \mathrm{KI}+8 \mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow 6 \mathrm{K}_{2} \mathrm{SO}_{4}+2 \mathrm{MnSO}_{4}+5 \mathrm{I}_{2}+8 \mathrm{H}_{2} \mathrm{O} \end{array}$$

In your own words, define or explain these terms or symbols. (a) \(\stackrel{\Delta}{\longrightarrow}\) (b) (aq) (c) stoichiometric coefficient (d) overall equation
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