Answer:
0.5
Explanation:
We are given the moles of two reactants, so this could be a limiting reactant problem.
We know that we will need moles, so, lets assemble all the data in one place.
2Ca₃(PO₄)₂ + 6SiO₂ + 10C → P₄ + 6CaSiO₃ + 10CO
n/mol: 1 3
Calculate the moles of P₄ that can be formed from each reactant :
1. From Ca₃(PO₄)₂
[tex]\text{Moles of P}_{4} = \text{1 mol Ca$_{3}$(PO}_{4})_{2} \times \dfrac{\text{1 mol P}_{4}}{\text{2 mol Ca$_{3}$(PO}_{4})}_{2} = \text{0.5 mol P}_{4}[/tex]
2. From SiO₂
[tex]\text{Moles of P}_{4} = \text{3 mol SiO}_{2} \times \dfrac{\text{1 mol P}_{4}}{\text{6 mol SiO}_{2}} = \text{0.5 mol P}_{4}[/tex]
Each reactant forms 0.5 mol of P₄.
