Question

The use of silica to form slag in the production of phosphorus from phosphate rock was introduced by Robert Boyle more than 300 years ago. When fluorapatite \(\left[\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{~F}\right]\) is used in phosphorus production, most of the fluorine atoms appear in the slag, but some end up in the toxic and corrosive gas \(\mathrm{SiF}_{4}\) (a) If \(15 \%\) by mass of the fluorine in \(100 . \mathrm{kg}\) of \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{~F}\) forms SiF \(_{4}\) what volume of this gas is collected at 1.00 atm and the industrial furnace temperature of \(1450 .{ }^{\circ} \mathrm{C} ?\) (b) In some facilities, the \(\mathrm{SiF}_{4}\), is used to produce sodium hexafluorosilicate \(\left(\mathrm{Na}_{2} \mathrm{SiF}_{6}\right),\) which is sold for water fluoridation: \(2 \mathrm{SiF}_{4}(g)+\mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l)\) $$ \mathrm{Na}_{2} \mathrm{SiF}_{6}(a q)+\mathrm{SiO}_{2}(s)+\mathrm{CO}_{2}(g)+2 \mathrm{HF}(a q) $$ How many cubic meters of drinking water can be fluoridated to a level of 1.0 ppm of \(F\) using the \(\mathrm{SiF}_{4}\) produced in part (a)?

Short answer

310 cubic meters of drinking water can be fluoridated.

Step-by-step solution

Identify the amount of fluorine in the initial mass

Given 100 kg of \(\text{Ca}_{5}(\text{PO}_{4})_{3} \text{F},\) find the mass of fluorine. The molar mass of \(\text{Ca}_{5}(\text{PO}_{4})_{3}\text{F}\) is \[5\times40.08 + 3\times(2\times 30.97 + 4\times16) + 19 = 504.31\text{{ g/mol}}\] Fluorine contributes 19 g per 504.31 g. The fraction of fluorine by mass is \(\frac{19}{504.31}\text{{ kg}}\). Therefore, total fluorine mass in 100 kg is \[ \text{mass}_F = \frac{19}{504.31} \times 100 = 3.77\text{{ kg}}\]

Calculate the mass of fluorine forming SiF\textsubscript{4}

15% of fluorine forms SiF\textsubscript{4}. Thus: \[ \text{mass}_{\text{F}}^{\text{SiF}_4} = 0.15 \times 3.77 = 0.566 \text{{ kg}} \]

Find the moles of SiF\textsubscript{4} produced

Molar mass of SiF\textsubscript{4} is \[ 28.09 (\text{{Si}}) + 4 \times 19 (\text{{F}}) = 104.09 \text{{ g/mol}} \] Moles of SiF\textsubscript{4} are \[ \text{moles SiF}_4 = \frac{0.566 \times 10^3}{104.09} = 5.44 \text{{ mol}}\]

Use the ideal gas law to find the volume of SiF\textsubscript{4}

Using ideal gas law \[ PV = nRT \] With \(P = 1.00 \text{ atm}, n = 5.44 \text{ mol}, T = 1450 + 273 = 1723 \text{ K}, R = 0.0821 \text{ L·atm/K·mol}\), \[ V = \frac{nRT}{P} = \frac{5.44 \times 0.0821 \times 1723}{1.00} = 768 \text{ L}\]

Determine the amount of Na\textsubscript{2}SiF\textsubscript{6} produced

In reaction \[2 \text{SiF}_4 + \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} \rightarrow \text{Na}_2\text{SiF}_6(aq) + \text{SiO}_2(s) + \text{CO}_2(g) + 2 \text{HF}(aq)\] According to stoichiometry, 2 moles of SiF\textsubscript{4} produce 1 mole Na\textsubscript{2}SiF\textsubscript{6}. Thus, 5.44 moles SiF\textsubscript{4} produce \[ \frac{5.44}{2} = 2.72 \text{{ moles Na}}_2\text{{SiF}}_6 \]

Calculate how much fluorine is in Na\textsubscript{2}SiF\textsubscript{6}

Each mole of Na\textsubscript{2}SiF\textsubscript{6} contains 6 atoms of fluorine. Therefore, total moles of fluorine = \[6 \times 2.72 = 16.32 \text{{ mol}} . \]

Calculate the mass of fluorine

Mass of fluorine: \[ \text{{mass}} = 16.32 \times 19 \text{{ g/mol}} = 310.08 \text{{ g}} = 0.31008 \text{{ kg}}\]

Determine the volume of drinking water fluoridated to 1.0 ppm

1 ppm means 1 g of fluorine in 1,000,000 g of water, which is 1 mg/L. So, 0.31008 kg of fluorine fluoridates: \[ \text{volume}_{H_2O} = \frac{0.31008 \times 10^6}{1} = 310,080 \text{ L} = 310 \text{ m}^3 \]

Key concepts

fluorapatite

Fluorapatite is a naturally occurring mineral with the formula \(\text{Ca}_{5}(\text{PO}_{4})_{3} \text{F}\). It is a key component of phosphate rocks, which are used extensively in the production of phosphorus. Phosphorus is vital for various industrial applications, including fertilizers, chemicals, and even in the production of certain types of glass.
Fluoapatite is mainly composed of calcium, phosphorus, and fluorine. The fluorine content in fluorapatite can have significant industrial implications, as it often escapes as silicon tetrafluoride (SiF₄), a toxic gas formed during the processing of phosphate rock.
Understanding fluorapatite's composition helps in calculating the amount of fluorine available for chemical reactions, essential for both producing desired products and controlling harmful emissions.

ideal gas law

The Ideal Gas Law is an equation of state for a hypothetical ideal gas, represented as \(PV = nRT\). Here:

• P stands for pressure
• V stands for volume
• n stands for the number of moles
• R is the universal gas constant, typically 0.0821 L·atm/K·mol
• T is the temperature in Kelvin.

In the given problem, it is used to find the volume of SiF₄ produced at 1.00 atm and a temperature of 1450°C (which is 1723 K). By applying the ideal gas law, the volume can be effortlessly calculated, ensuring that chemical engineers can predict and manage the amounts of gases produced in industrial processes.

stoichiometry

Stoichiometry involves quantitative relationships between reactants and products in a chemical reaction. It’s crucial for determining how much of a substance is consumed or produced.
In the exercise, stoichiometry helps us determine the quantities of \(SiF_{4}\) and \(Na_{2}SiF_{6}\) formed. For instance, the balanced equation \(2 SiF_{4}+ Na_{2}CO_{3} + H_{2}O \rightarrow Na_{2}SiF_{6} + SiO_{2} + CO_{2} + 2 HF\) indicates that 2 moles of \(SiF_{4}\) yield 1 mole of \(Na_{2}SiF_{6}\). Knowing these ratios, one can compute the specific amounts formed during the reactions.

silicon tetrafluoride

Silicon tetrafluoride (SiF₄) is a toxic and corrosive gas. Its formation occurs during the treatment of fluorapatite and other fluoride-containing materials with acids or high temperatures. SiF₄ production is significant in this context as it not only poses environmental hazards but also serves as a precursor for useful industrial chemicals.
Its production, handling, and subsequent reactions (like forming \(Na_{2}SiF_{6}\)) are closely managed to balance industrial benefits with environmental safety. For instance, capturing and converting SiF₄ into less harmful compounds like \(Na_{2}SiF_{6}\) makes its presence in industrial applications more manageable.

drinking water fluoridation

Drinking water fluoridation involves adding fluoride compounds to water supplies to prevent tooth decay. Ensuring that fluoride levels stay within recommended limits (usually about 1.0 ppm) is critical for consumer safety.
In the exercise, the SiF₄ gas captured is transformed into sodium hexafluorosilicate (Na₂SiF₆). This compound is then used for fluoridating drinking water.
By determining how much fluorine is present in Na₂SiF₆, it’s possible to calculate the volume of water that can be fluoridated to safe levels. For instance, 0.31008 kg of fluorine can adequately fluoridate 310 cubic meters of drinking water, ensuring safe and beneficial levels of fluoride for public health.