White phosphorus is prepared by heating phosphate rock [principally \(\left.\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\right]\) with sand and coke: \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+\mathrm{SiO}_{2}(s)+\mathrm{C}(s) \longrightarrow$$$ \mathrm{CaSiO}_{3}(s)+\mathrm{CO}(g)+\mathrm{P}_{4}(g)[\text { unbalanced }]$$ How many kilograms of phosphate rock are needed to produce \)315 \mathrm{~mol}\( of \)\mathrm{P}_{4},\( assuming that the conversion is \)90 . \%$ efficient?

In chemistry, it is crucial to use balanced chemical equations. They ensure that the law of conservation of mass is met by having the same number of each type of atom on both sides of the equation.
To balance a chemical equation, adjust the coefficients before each reactant and product until the number of atoms for each element is the same on both sides.
In our example, the unbalanced reaction is: \( \mathrm{Ca}_{3}(\mathrm{PO}_{4})_{2} + \mathrm{SiO}_{2} + \mathrm{C} \rightarrow \mathrm{CaSiO}_{3} + \mathrm{CO} + \mathrm{P}_{4} \).
After balancing, it becomes: \( 2\ \mathrm{Ca}_{3}(\mathrm{PO}_{4})_{2} + 6\ \mathrm{SiO}_{2} + 10\ \mathrm{C} \rightarrow 6\ \mathrm{CaSiO}_{3} + 10\ \mathrm{CO} + \mathrm{P}_{4} \).
This balanced equation helps us determine the stoichiometric relationships between reactants and products.

Molar mass is the mass of one mole of a substance (usually in g/mol). It's a critical concept for converting between moles and grams.
To find the molar mass of a compound, sum the molar masses of its constituent atoms. For molecule \( \mathrm{Ca}_{3}(\mathrm{PO}_{4})_{2} \), we have:

• Calcium (Ca): \(40 \ \mathrm{g/mol} \)
• Phosphorus (P): \(31 \ \mathrm{g/mol} \)
• Oxygen (O): \(16 \ \mathrm{g/mol} \)

Adding these up:
\(3 \times 40 + 2 \times (31 + 4 \times 16) = 310 \ \mathrm{g/mol} \).
Therefore, the molar mass of \( \mathrm{Ca}_{3}(\mathrm{PO}_{4})_{2} \) is 310 g/mol.
This allows us to convert moles of phosphate rock to grams when needed.

Reaction efficiency tells us how much of the reactants are turned into products. It is represented as a percentage.
If a reaction has 90% efficiency, it means 90% of the reactants effectively produce the desired product. To account for efficiency, divide the theoretical yield by the efficiency (expressed as a decimal).
For example:

• Calculated moles required: 630 moles of \( \mathrm{Ca}_{3}(\mathrm{PO}_{4})_{2} \)
• Conversion efficiency: 90%

Effective moles required: \( \frac{630}{0.9} = 700 \ \mathrm{moles} \).
We need 700 moles of phosphate rock to achieve the 315 moles of \( \mathrm{P}_{4} \) at 90% efficiency.

Converting between moles and mass is a fundamental skill in stoichiometry.
The relationship is:
\( \text{Mass} = \text{Moles} \times \text{Molar Mass} \).
This formula allows conversions between these two quantities.
For example, with 700 moles of \( \mathrm{Ca}_{3}(\mathrm{PO}_{4})_{2} \) and its molar mass being 310 g/mol, the mass is:

• Mass = 700 moles \( \times 310 \ \mathrm{g/mol} = 217,000\ \mathrm{g} \)

Converting grams to kilograms (since 1 kg = 1000 g):
\( 217,000\ \mathrm{g} = 217\ \mathrm{kg} \).
Thus, 217 kg of phosphate rock is needed.