Final answer:
To determine the maximum amount of P₄S₁₀ formed, first balance the chemical equation, calculate moles for each reactant, identify the limiting reagent, and use it to calculate the moles of P₄S₁₀ formed, finally convert to grams. The maximum yield of P₄S₁₀ is 20.45 g.
Step-by-step explanation:
The question involves stoichiometry, which is the part of chemistry that studies the quantitative relationships between the amounts of reactants and products in chemical reactions. To solve this, first, balance the chemical equation:
P₄ + 6S₈ → 4 P₄S₁₀
Next, calculate the moles of each reactant provided:
- P₄: Molar mass ≈ 123.9 g/mol. Therefore, moles of P₄ = 15.29 g / 123.9 g/mol ≈ 0.123 mol.
- S₈: Molar mass ≈ 256.5 g/mol. Therefore, moles of S₈ = 17.66 g / 256.5 g/mol ≈ 0.069 mol.
Using the balanced equation: 1 mol P₄ reacts with 6 mol S₈. The limiting reagent can be found, which in this case is S₈ since ≈ 0.069 mol is less than 6 * 0.123 mol (which is the amount of S₈ needed for the reaction to completely use up the P4).
Since S⁸ is the limiting reagent, use its moles to find the amount of P₄S₁₀produced:
4 moles of P₄S₁₀ are produced from 6 moles of S⁸, so:
Moles of P₄S₁₀ = ≈ 4/6 * 0.069 mol = 0.046 mol.
Finally, calculate the mass of P₄S₁₀ formed: Molar mass of P₄S₁₀ ≈ 444.6 g/mol.
Mass of P₄S₁₀ = 0.046 mol * 444.6 g/mol ≈ 20.45 g.
Therefore, the maximum amount of P₄S₁₀ that could be formed from the given amounts of reactants is 20.45 g, expressed to three significant figures as the problem instruction suggests.