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Chapter 4: Problem 71

Solid calcium fluoride (CaF, \(_{2}\) ) reacts with sulfuric acid to form solid calcium sulfate and gaseous hydrogen fluoride (HF). The HF is then dissolved in water to form hydrofluoric acid. A source of calcium fluoride is fluorite ore containing \(96.0 \mathrm{wt} \% \mathrm{CaF}_{2}\) and \(4.0 \% \mathrm{SiO}_{2}\) In a typical hydrofluoric acid manufacturing process, fluorite ore is reacted with 93 wt\% aqueous sulfuric acid, supplied 15\% in excess of the stoichiometric amount. Ninety-five percent of the ore dissolves in the acid. Some of the HF formed reacts with the dissolved silica in the reaction $$6 \mathrm{HF}+\mathrm{SiO}_{2}(\mathrm{aq}) \rightarrow \mathrm{H}_{2} \mathrm{SiF}_{6}(\mathrm{s})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})$$ The hydrogen fluoride exiting from the reactor is subsequently dissolved in enough water to produce 60.0 wt\% hydrofluoric acid. Calculate the quantity of fluorite ore needed to produce a metric ton of aqueous hydrofluoric acid. Note: Some of the given data are not needed to solve the problem.

Short Answer

Expert verified
Hence, to produce one metric ton of 60 wt% hydrofluoric acid, we need approximately 1.316 metric tons (or 1316 kilograms) of fluorite ore.

Step by step solution

01

Flouride Ore to Hydrogen Flouride

This problem starts with solid calcium fluoride (CaF\(_{2}\)) reacting with sulfuric acid to form hydrogen fluoride (HF). This part of the reaction is described by the balanced equation: CaF\(_{2}\) + H\(_{2}\)SO\(_{4}\) \(\rightarrow\) 2HF + CaSO\(_{4}\). This means one mole of calcium fluoride reacts to form two moles of hydrogen fluoride. In terms of weight, 78.08 g of CaF2 will react to form 40.02g of HF (close to a 1:0.5 ratio).
02

Weight of HF to be produced

One metric ton (1000 kg) of 60 wt% aqueous HF needs to be produced. This means the total amount of HF needed is 1000kg*0.60 = 600kg. Considering the 1:0.5 ratio, to produce 600 kg HF, around 1200 kg (600kg/0.5) of CaF\(_2}\) is needed.
03

Reactor Efficiency

However, in the manufacturing process, not all of the fluorite ore (CaF\(_2}\)) reacts with sulfuric acid. Only 95% of the ore dissolves in acid. This means the amount of flourite ore needed would be higher: 1200kg/95% = 1263.16 kg or 1.263 metric tons.
04

Composition of fluorite ore

Fluorite ore has only 96 wt% calcium fluoride (the rest is irrelevant silica). So we also need to take into account the actual CaF\(_{2}\) content in the ore. So, to get 1.263 tons of pure CaF\(_2}\), we need to use more fluorite ore, which is given by: 1.263 tons / 0.96 = 1.31563 tons

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Fluorite Ore
Fluorite ore, also known as fluorspar, is a mineral primarily made up of calcium fluoride ( \( \text{CaF}_2 \) ). In the context of chemical reactions, particularly in the production of hydrogen fluoride, fluorite ore serves as a crucial reactant. It is commonly extracted from mining sites where it is found alongside various impurities.

One of the impurities found in fluorite ore can be silica ( \( \text{SiO}_2 \) ). This impurity does not contribute to the desired chemical reaction but can interact with the produced hydrogen fluoride, as shown in the given reaction formula. In industrial applications, the purity of fluorite ore is often considered, with typical ore containing around 96% \( \text{CaF}_2 \).

To calculate the amount of fluorite ore needed for any chemical production, it's essential to consider the ore's purity. This helps to determine how much actual reactant is available in a given sample of ore. In the case of the given exercise, knowing the ore contains 96% \( \text{CaF}_2 \) lets you adjust your quantities accordingly to ensure an efficient production process.
Hydrogen Fluoride Production Process
Hydrogen fluoride (HF) is a valuable chemical, primarily used in the production of hydrofluoric acid. The production process involves a chemical reaction where calcium fluoride from fluorite ore interacts with sulfuric acid.

The balanced chemical equation describes the reaction as:\[\text{CaF}_2 + \text{H}_2\text{SO}_4 \rightarrow 2 \text{HF} + \text{CaSO}_4\]This reaction shows that one mole of calcium fluoride ( \( \text{CaF}_2 \) ) reacts with one mole of sulfuric acid to produce two moles of hydrogen fluoride and one mole of calcium sulfate as a byproduct.

In industrial applications, sulfuric acid is usually supplied in excess to increase reaction efficiency. In this particular exercise, the sulfuric acid is supplied at 15% excess, ensuring nearly complete conversion of \( \text{CaF}_2 \) to hydrogen fluoride. However, only 95% of the ore dissolves in practice, which is an important consideration when calculating the required initial amount of fluorite ore.
How to Calculate Reactant Quantities
Calculating the correct quantities of reactants is fundamental in stoichiometry. For hydrogen fluoride production, it entails determining how much fluorite ore is needed to obtain a desired amount of hydrogen fluoride.

In the exercise, you want to produce 600 kg of hydrogen fluoride, which translates to a total requirement of 1000 kg of hydrofluoric acid at a 60% concentration. First, calculate the moles of HF needed by using the molar mass of HF and the desired mass.

Given the 1:0.5 reaction ratio from the balanced equation, double the moles of HF required to find the moles of \( \text{CaF}_2 \). Convert these moles to a mass using the molar mass of \( \text{CaF}_2 \), accounting for the 95% reaction efficiency, which means that only part of the fluorite ore reacts effectively.
  • Begin by adjusting your calculations to account for the purity of the fluorite ore (96% purity).
  • Use these adjustments to find the mass of ore needed to provide the pure \( \text{CaF}_2 \).
  • The calculations will lead you from the desired amount of hydrofluoric acid back to the initial mass of fluorite ore required.
Understanding these calculations helps ensure that the chemical reaction has enough reactants to proceed efficiently without unnecessary excess, which saves costs and resources in industrial practice.

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Most popular questions from this chapter

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