# Question e4dc6

Nov 3, 2015

Here's what I got.

#### Explanation:

The important thing to do here is write a balanced chemical equation for this decomposition reaction.

Sodium bicarbonate, ${\text{NaHCO}}_{3}$, will decompose to form sodium carbonate, ${\text{Na"_2"CO}}_{3}$, water, and carbon dioxide, ${\text{CO}}_{2}$

$\textcolor{red}{2} {\text{NaHCO"_text(3(s]) -> "Na"_2"CO"_text(3(s]) + "CO"_text(2(g]) + "H"_2"O}}_{\textrm{\left(g\right]}}$

Notice that you have a $\textcolor{red}{2} : 1$ mole ratio between sodium bicarbonate and sodium carbonate. This means that the reaction will produce half as many moles of the latter than whatever number of moles of the former underwent decomposition.

Use sodium carbonate's molar amss to determine how many moles you'd get in that sample

0.685color(red)(cancel(color(black)("g"))) * "1 mole NaHCO"_3/(84.007color(red)(cancel(color(black)("g")))) = "0.008154 moles NaHCO"_3

Now, if the reaction were to have a 100% yield, it would produce

0.008154color(red)(cancel(color(black)("moles NaHCO"_3))) * ("1 mole Na"_2"CO"_3)/(color(red)(2)color(red)(cancel(color(black)("moles NaHCO"_3)))) = "0.004077 moles Na"_2"CO"_3

Use the molar mass of sodium carbonate to determine how many grams would contain this many moles

0.004077color(red)(cancel(color(black)("moles"))) * "105.99 g"/(1color(red)(cancel(color(black)("mole")))) = "0.4321 g Na"_2"CO"_3

This will be your reaction's theoretical yield, which tells you how much product you can expect when all the moles of the reactant actually form product.

Now, you know that the reaction produced $\text{0.418 g}$, a little less than what you expected for a 100% yield reaction. This is your actual yield.

This tells you that not all the moles of sodium carbonate reacted to produce sodium carbonate. In other words, the reaction did not have a 100% yield.

Percent yield is defined as

$\textcolor{b l u e}{\text{% yield" = "actual yield"/"theoretical yield} \times 100}$

"% yield" = (0.418color(red)(cancel(color(black)("g"))))/(0.4321color(red)(cancel(color(black)("g")))) xx 100 = color(green)(96.7%)#