Answer:
0.5
Step-by-step 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
![\text{Moles of P}_(4) = \text{1 mol Ca$_(3)$(PO}_(4))_(2) * \frac{\text{1 mol P}_(4)}{\text{2 mol Ca$_(3)$(PO}_(4))}_(2) = \text{0.5 mol P}_(4)](https://img.qammunity.org/2021/formulas/chemistry/high-school/j3kqytuvkzwejfbz2eh37lronpwfj0rv5x.png)
2. From SiO₂
![\text{Moles of P}_(4) = \text{3 mol SiO}_(2) * \frac{\text{1 mol P}_(4)}{\text{6 mol SiO}_(2)} = \text{0.5 mol P}_(4)](https://img.qammunity.org/2021/formulas/chemistry/high-school/l3pnspcjdmj57n48def36yf7p6jl4eekud.png)
Each reactant forms 0.5 mol of P₄.