Chapter U4, Problem 19STP

Interpretation Introduction

Interpretation:

The number grams of Zn (PO₄)₂ can be made with 26.5 g of Zn (NO₃)₂ and 19.7 g of Na₃PO₄, needs to be identified.

Concept introduction:

The limiting reagents are also known as limiting reactants, in which the compounds from the reactant are completely consumed during the reaction.

Answer to Problem 19STP

The correct answer is (b).

Explanation of Solution

The given reaction is as follows:

$\text{3Zn}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\left(\text{aq}\right){\text{ + 2Na}}_{\text{3}}{\text{PO}}_{\text{4}}\left(\text{aq}\right)\to {\text{Zn}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\left(\text{s}\right){\text{ + 6NaNO}}_{\text{3}}\left(\text{aq}\right)$

${\text{moles of Zn(NO}}_3{)}_2=\frac{\text{mass}}{\text{molar mass}}=\frac{\text{26}\text{.5g}}{189.36 \text{g/mol}}=0.14\text{mol}$

Amount of Na₃PO₄ = 19.7 g

${\text{moles of Na}}_3{\text{PO}}_{\text{4}}=\frac{\text{mass}}{\text{molar mass}}=\frac{\text{19}\text{.7g}}{163.94 \text{g/mol}}=0.12\text{mol }$

Here, 2 mol of Na₃(PO₄)₂ required 3 mol of Zn (NO₃)₂ thus 0.12 mol of Na₃(PO₄)₂ required:

$\text{3}\times \frac{\text{0}\text{.12}}{2}\text{= 0}\text{.18 mol of Zn}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$

So, the limiting reagent is Zn (NO₃)_2.

3 moles of Zn (NO₃)₂produces 1 mole of Zn₃(PO₄)₂:

So, the mass of Zn₃(PO₄)₂ = $\frac{\text{mass }\times \text{ molar mass}}{\text{2}}$

= $\frac{0.14\times 386.11}{3}=18.01\approx 18\text{ g}$

Thus, the correct answer is (b).