# Question 02fad

Jun 29, 2015

The excess reagent is lithium phosphate.

#### Explanation:

To determine which reactant is in excess, take a look at the balanced chemical equation for this double replacement reaction

$\textcolor{red}{3} C a {\left(N {O}_{3}\right)}_{2 \left(a q\right)} + \textcolor{g r e e n}{2} L {i}_{3} P {O}_{4 \left(a q\right)} \to 6 L i N {O}_{3 \left(a q\right)} + C {a}_{3} {\left(P {O}_{4}\right)}_{2 \left(s\right)} \downarrow$

Notice that you have a $\textcolor{red}{3} : \textcolor{g r e e n}{2}$ mole ratio between calcium nitrate and lithium phosphate. This means that the moles of each reactant that react will always respect this ratio.

So, you know that you have 3.4 moles of calcium nitrate and 2.4 moles of lithium phosphate. To see whether or not you'rea dealing with a limiting reagent, check if the mole ratio is respected

3.4cancel("moles"Ca(NO_3)_2) * (color(green)(2)" mole "Li_3PO_4)/(color(red)(3)cancel("moles"Ca(NO_3)_2)) = "2.27 moles" $L {i}_{3} P {O}_{4}$

This is less than the number of moles available for lithium phosphate, which is 2.4. This means that you have insufficient calcium nitrate.

More specifically, you'd need

2.4cancel("moles"Li_3PO_4) * (color(red)(3)" moles "Ca(NO_3)_2)/(color(green)(2)cancel("moles"Li_3PO_4)) = "3.6 moles"# $C a {\left(N {O}_{3}\right)}_{2}$

in order to react completely with 2.4 moles of lithium phosphate.

Therefore, lithium phosphate is in excess and calcium nitrate is the limiting reagent.

You have an excess of

${n}_{\text{excess" = underbrace("2.4 moles")_(color(blue)("what you have")) - overbrace("2.27 moles")^(color(red)("what you use")) = "0.13 moles}}$ $L {i}_{3} P {O}_{4}$