# Question #05456

Aug 15, 2017

WARNING! Long answer! The reaction will produce 370 g of ${\text{Na"_2"S"_2"O}}_{3}$.

#### Explanation:

We are given the amounts of three reactants, so this is a limiting reactant problem.

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

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

${M}_{r} : \textcolor{w h i t e}{m m m m m} 78.04 \textcolor{w h i t e}{m m} 105.99 \textcolor{w h i t e}{m m} 64.06 \textcolor{w h i t e}{m m l l} 158.11$
$\textcolor{w h i t e}{m m m m m m m} {\text{2Na"_2"S" + "Na"_2"CO"_3 + "4SO"_2 → "3Na"_2"S"_2"O"_3 + "CO}}_{2}$
$\text{Mass/g} : \textcolor{w h i t e}{m m m l} 200 \textcolor{w h i t e}{m m m l l} 200 \textcolor{w h i t e}{m m l l} 200$
$\text{Amt/mol:} \textcolor{w h i t e}{m m l} 2.563 \textcolor{w h i t e}{m m l l} 1.887 \textcolor{w h i t e}{m m} 3.122$
$\text{Divide by:} \textcolor{w h i t e}{m m m} 2 \textcolor{w h i t e}{m m m m m} 1 \textcolor{w h i t e}{m m m m} 4$
$\text{Moles rxn:} \textcolor{w h i t e}{m m} 1.281 \textcolor{w h i t e}{m m l l} 1.887 \textcolor{w h i t e}{m l l} 0.7805$

$\text{Moles of Na"_2"S" = 200 color(red)(cancel(color(black)("g Na"_2"S"))) × ("1 mol Na"_2"S")/(78.04 color(red)(cancel(color(black)("g Na"_2"S")))) = "2.563 mol Na"_2"S}$

${\text{Moles of Na"_2"CO"_3 =200 color(red)(cancel(color(black)("g Na"_2"CO"_3))) × ("1 mol Na"_2"CO"_3)/(105.99 color(red)(cancel(color(black)("g Na"_2"CO"_3)))) = "1.887 mol Na"_2"CO}}_{3}$

${\text{Moles of SO"_2 = 200 color(red)(cancel(color(black)("g SO"_2))) × ("1 mol SO"_2)/(64.06 color(red)(cancel(color(black)("g SO"_2)))) = "0.7805 mol SO}}_{2}$

2. Identify the limiting reactant

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.

${\text{SO}}_{2}$ is the limiting reactant because it gives the fewest moles of reaction.

3. Calculate the theoretical moles of ${\text{Na"_2"S"_2"O}}_{3}$

$\text{Theoretical moles" = 3.122 color(red)(cancel(color(black)("mol SO"_2))) × ("3 mol Na"_2"S"_2"O"_3)/(4 color(red)(cancel(color(black)("mol SO"_2)))) = "2.342 mol Na"_2"S"_2"O"_3}$

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

${\text{Theoretical yield" = 2.342 color(red)(cancel(color(black)("mol Na"_2"S"_2"O"_3))) × ("158.11 g Na"_2"S"_2"O"_3)/(1 color(red)(cancel(color(black)("mol Na"_2"S"_2"O"_3)))) = "370 g Na"_2"S"_2"O}}_{3}$