Question

Please solve problem 2 in the attachment

Answer 1

Answer: 0.16 moles of $M_3PO_4$ 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:

$6MX+L_3(PO_4)_2\rightarrow 2M_3PO_4+3LX_2$

To calculate limiting reagent, we need to compare the actual ratio of moles to the stoichiometric ratio of moles of the reactants.

$\text{Actual Ratio}=\frac{\text{0.2 moles of }L_3(PO_4)_2}{\text{0.5 moles of MX}}$

$\text{Actual Ratio}=\frac{\text{0.4 moles of }L_3(PO_4)_2}{\text{1 mole of MX}}$

Now, the stoichiometric ratio:

$\text{Stoichiometric Ratio}=\frac{\text{1 mole of }L_3(PO_4)_2}{\text{6 moles of MX}}$

$\text{Stoichiometric Ratio}=\frac{\text{0.16 moles of }L_3(PO_4)_2}{\text{1 mole of MX}}$

This means that at least 0.16 moles of $L_3(PO_4)_2$ is required for every 1 mole of MX. As, the actual ratio is greater than the stoichiometric ratio, so $L_3(PO_4)_2$ 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 $M_3PO_4$

So, 0.5 moles of MX will produce = $\frac{2}{6}\times 0.5\text{ moles of }M_3PO_4$

                                                       = $\text{0.16 moles of }M_3PO_4$

Hence, the correct option is C.