Question

QUESTION 1. Consider the reaction . MnO2 + 4 HCl → MnCl2 + Cl2 + 2H2O
If 0.86 mole of MnO2 and 48.2 g of HCl react, which reagent will be used up first? How many
grams of Cl2 will be produced?

QUESTION 2 .Titanium(VI) oxide (TiO2) is a white substance produced by the action of sulfuric acid on the
mineral ilmenite (FeTiO3):
FeTiO3 + H2SO4 → TiO2 + FeSO4 + H2O
Its opaque and nontoxic properties make it suitable as a pigment in plastics and paints. In one
process, 8.00 x 103 kg of FeTiO3 yielded 3.67 x 103 kg of TiO2. What is the percent yield of the
reaction?

Answer 1

Answer on Question #53168 – Chemistry – Inorganic Chemistry

Question 1:

Consider the reaction: $\mathrm{MnO}_2 + 4\mathrm{HCl} \rightarrow \mathrm{MnCl}_2 + \mathrm{Cl}_2 + 2\mathrm{H}_2\mathrm{O}$

If 0.86 mole of $\mathrm{MnO}_2$ and $48.2\ \mathrm{g}$ of HCl react, which reagent will be used up first? How many grams of $\mathrm{Cl}_2$ will be produced?

Answer 1:

Given

$$\begin{array}{l} n(\mathrm{MnO}_2) = 0.86\ \mathrm{mole} \\ m(\mathrm{HCl}) = 48.2\ \mathrm{g} \\ \mathrm{M}(\mathrm{HCl}) = 36.5\ \mathrm{g}/\mathrm{mole},\ \mathrm{M}(\mathrm{Cl}_2) = 71\ \mathrm{g}/\mathrm{mole} \\ \end{array}$$ $$n(\mathrm{HCl}) = m(\mathrm{HCl})/\mathrm{M}(\mathrm{HCl}) = 48.2 / 36.5 = 1.32\ \mathrm{mole}$$

From the chemical equation: 1 mole of $\mathrm{MnO}_2$ reacts with 4 moles of HCl.

Thus, 0.86 mole of $\mathrm{MnO}_2$ requires $4 \cdot 0.86 = 3.44$ moles of HCl to react with (but only 1.32 mole are present). This means there is deficiency of HCl in terms of amount of $\mathrm{MnO}_2$, and HCl will be used up first.

From the chemical equation: 1 mole of $\mathrm{Cl}_2$ is produced when 4 moles of HCl react. So, we can write the following proportion:

$$\begin{array}{l} 4\ \mathrm{mole} \cdot 36.5\ \mathrm{g}/\mathrm{mole}(\mathrm{HCl}) - 1\ \mathrm{mole} \cdot 71\ \mathrm{g}/\mathrm{mole}(\mathrm{Cl}_2) \\ 48.2\ \mathrm{g}(\mathrm{HCl}) - m(\mathrm{Cl}_2) \\ \end{array}$$ $$m(\mathrm{Cl}_2) = 71 \cdot 48.2 / (4 \cdot 36.5) = 23.4\ \mathrm{g}$$

Answer: HCl will be used up first.

23.4 g of $\mathrm{Cl}_2$ will be produced.

Question 2:

Titanium(VI) oxide $(\mathrm{TiO}_2)$ is a white substance produced by the action of sulfuric acid on the mineral ilmenite $(\mathrm{FeTiO}_3)$:

$$\mathrm{FeTiO}_3 + \mathrm{H}_2\mathrm{SO}_4 \rightarrow \mathrm{TiO}_2 + \mathrm{FeSO}_4 + \mathrm{H}_2\mathrm{O}$$

Its opaque and nontoxic properties make it suitable as a pigment in plastics and paints. In one process, $8.00 \cdot 10^3\ \mathrm{kg}$ of $\mathrm{FeTiO}_3$ yielded $3.67 \cdot 10^3\ \mathrm{kg}$ of $\mathrm{TiO}_2$. What is the percent yield of the reaction?

Answer 2:

Given

$$\begin{array}{l} m(\mathrm{FeTiO}_3) = 8.00 \cdot 10^3\ \mathrm{kg} \\ m(\mathrm{TiO}_2) = 3.67 \cdot 10^3\ \mathrm{kg} \\ \mathrm{M}(\mathrm{FeTiO}_3) = 151.71\ \mathrm{g}/\mathrm{mole},\ \mathrm{M}(\mathrm{TiO}_2) = 79.87\ \mathrm{g}/\mathrm{mole} \\ \end{array}$$

As is clear from the chemical equation, 1 mole of $\mathrm{TiO}_2$ is produced when 1 mole of $\mathrm{FeTiO}_3$ react.

So, we can write the following proportion:

$$\begin{array}{l} 151.71\ \mathrm{g}/\mathrm{mole}(\mathrm{FeTiO}_3) - 79.87\ \mathrm{g}/\mathrm{mole}(\mathrm{TiO}_2) \\ 8.00 \cdot 10^3\ \mathrm{kg}(\mathrm{FeTiO}_3) - m(\mathrm{TiO}_2)(\text{theoretical}) \\ \end{array}$$ $$\begin{array}{l} m(\mathrm{TiO}_2)(\text{theoretical}) = 79.87 \cdot 8.00 \cdot 10^3 / 151.71 = 4.21 \cdot 10^3\ \mathrm{kg} \\ \% \text{Yield}(\mathrm{TiO}_2) = m(\mathrm{TiO}_2) / m(\mathrm{TiO}_2)(\text{theoretical}) \cdot 100\% = 3.67 \cdot 10^3 / 4.21 \cdot 10^3 \cdot 100\% = 87.17\% \\ \end{array}$$

Answer: $\% \text{Yield}(\mathrm{TiO}_2) = 87.17\%$

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