# Question 751ec

Aug 17, 2017

ΔH^@ = "-219 kJ/mol"

#### Explanation:

This is a two-part question.

We are given the amounts of two reactants, so we must first identify the limiting reactant. Then we can calculate the value of ΔH^@.

Part 1. Identify the limiting reactant

We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.

Gather all the information in one place with molar masses above the formulas and everything else below the formulas.

${M}_{r} : \textcolor{w h i t e}{m m m m l} 207.2$
$\textcolor{w h i t e}{m m m m m m m} \text{2Pb" color(white)(ml)+color(white)(ml) "O"_2 → "2PbO}$
$\text{Mass/g:} \textcolor{w h i t e}{m m m} 34.1$
$\text{Moles:"color(white)(mmm)"0.1646"color(white)(mmm)"0.2782}$
$\text{Divide by:} \textcolor{w h i t e}{m m m} 2 \textcolor{w h i t e}{m m m m m} 1$
$\text{Moles rxn:"color(white)(m)"0.082 29"color(white)(mm)"0.2782}$

$\text{Moles of Pb" = 34.1 color(red)(cancel(color(black)("g Pb"))) × ("1 mol Pb")/(207.2 color(red)(cancel(color(black)("g Pb")))) = "0.1646 mol Pb}$

n_text(O₂) = (pV)/(RT) = (1.00 color(red)(cancel(color(black)("atm"))) × 6.81 color(red)(cancel(color(black)("L"))))/(0.0821 color(red)(cancel(color(black)("L·atm·K"^"-1")))"mol"^"-1" × 298.15 color(red)(cancel(color(black)("K")))) = "0.2782 mol"

An easy way to identify the limiting reactant is to calculate the "moles of reaction" each will give:

You divide the moles of each reactant by its corresponding coefficient in the balanced equation.

I did that for you in the table above.

$\text{Pb}$ is the limiting reactant because it gives fewer moles of reaction.

Part 2. Calculate ΔH^@

nΔ_text(rxn)H = "-36.1 kJ"

Δ_text(rxn)H = ("-36.1 kJ")/("0.1646 mol") = "-219 kJ/mol"#