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.

Step-by-step 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 =
`(2)/(6)* 0.5\text{ moles of }M_3PO_4`

=
`\text{0.16 moles of }M_3PO_4`

Hence, the correct option is C.