Answer: 0.16 moles of [tex]M_3PO_4[/tex] will be formed. Correct option is C.
Explanation: We are given 0.5 moles of Metal (I) halide and 0.2 moles of metal (II) phosphate. The reaction follows:
[tex]6MX+L_3(PO_4)_2\rightarrow 2M_3PO_4+3LX_2[/tex]
To calculate limiting reagent, we need to compare the actual ratio of moles to the stoichiometric ratio of moles of the reactants.
[tex]\text{Actual Ratio}=\frac{\text{0.2 moles of }L_3(PO_4)_2}{\text{0.5 moles of MX}}[/tex]
[tex]\text{Actual Ratio}=\frac{\text{0.4 moles of }L_3(PO_4)_2}{\text{1 mole of MX}}[/tex]
Now, the stoichiometric ratio:
[tex]\text{Stoichiometric Ratio}=\frac{\text{1 mole of }L_3(PO_4)_2}{\text{6 moles of MX}}[/tex]
[tex]\text{Stoichiometric Ratio}=\frac{\text{0.16 moles of }L_3(PO_4)_2}{\text{1 mole of MX}}[/tex]
This means that at least 0.16 moles of [tex]L_3(PO_4)_2[/tex] is required for every 1 mole of MX. As, the actual ratio is greater than the stoichiometric ratio, so [tex]L_3(PO_4)_2[/tex] is present in greater amount. Therefore, MX is considered as the limiting reagent because it limits the formation of product.
By Stoichiometry,
6 moles of MX produces 2 moles of [tex]M_3PO_4[/tex]
So, 0.5 moles of MX will produce = [tex]\frac{2}{6}\times 0.5\text{ moles of }M_3PO_4[/tex]
= [tex]\text{0.16 moles of }M_3PO_4[/tex]
Hence, the correct option is C.
