# Question #6d399

Jan 26, 2017

WARNING! Long answer! The limiting reactant is $\text{NaOH}$. The theoretical yield of ${\text{Na"_2"SO}}_{4}$ is 10.83 g. The mass of excess ${\text{H"_2"SO}}_{4}$ is 2.15 g.

#### Explanation:

This is a limiting reactant problem.

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

$\textcolor{b l u e}{\boldsymbol{\text{A."color(white)(l) "Identify the limiting reactant}}}$

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

${M}_{r} : \textcolor{w h i t e}{m m m m m} 98.08 \textcolor{w h i t e}{m m l l} 40.00 \textcolor{w h i t e}{m m l l} 142.04$
$\textcolor{w h i t e}{m m m m m m m} \text{H"_2"SO"_4 + "2NaOH" → "Na"_2"SO"_4 + "2H"_2"O}$
$\text{Mass/g:} \textcolor{w h i t e}{m m m l l} 9.65 \textcolor{w h i t e}{m m m} 6.10$
$\text{Amt/mol:"color(white)(ml)"0.098 38} \textcolor{w h i t e}{m l l} 0.1525$
$\text{Divide by:} \textcolor{w h i t e}{m m m} 1 \textcolor{w h i t e}{m m m m l l} 2$
$\text{Moles rxn:"color(white)(m)"0.098 38"color(white)(mll)"0.076 25}$

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.

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

$\textcolor{b l u e}{{\boldsymbol{\text{B."color(white)(l) "Calculate the theoretical yield of Na"_2"SO}}}_{4}}$

2. Calculate the theoretical moles of ${\text{Na"_2"SO}}_{4}$

${\text{Theoretical yield" = 0.1525 color(red)(cancel(color(black)("mol NaOH"))) × ("1 mol Na"_2"SO"_4)/(2 color(red)(cancel(color(black)("mol NaOH")))) = "0.076 25 mol Na"_2"SO}}_{4}$

3. Calculate the theoretical yield of ${\text{Na"_2"SO}}_{4}$

${\text{Theoretical yield" = "0.076 25" color(red)(cancel(color(black)("mol Na"_2"SO"_4))) × ("142.04 g Na"_2"SO"_4)/(1 color(red)(cancel(color(black)("mol Na"_2"SO"_4)))) = "10.83 g Na"_2"SO}}_{4}$

$\textcolor{b l u e}{\boldsymbol{\text{C."color(white)(l) "Calculate the mass of excess reagent}}}$

4. Calculate the moles of ${\text{H"_2"SO}}_{4}$ reacted.

${\text{Moles of H"_2"SO"_4 = 0.1525 color(red)(cancel(color(black)("mol NaOH"))) × ("1 mol H"_2"SO"_4)/(2 color(red)(cancel(color(black)("mol NaOH")))) = "0.076 25 mol H"_2"SO}}_{4}$

5. Calculate the mass of ${\text{H"_2"SO}}_{4}$ reacted.

${\text{Mass of H"_2"SO"_4 = "0.076 25" color(red)(cancel(color(black)("mol H"_2"SO"_4))) × ("98.08 g H"_2"SO"_4)/(1 color(red)(cancel(color(black)("mol H"_2"SO"_4)))) = "7.479 g H"_2"SO}}_{4}$

6. Calculate the mass of ${\text{H"_2"SO}}_{4}$ remaining.

${\text{Mass of H"_2"SO"_4 = "9.65 g - 7.479 g" = "2.15 g H"_2"SO}}_{4}$