# Question b5362

Nov 29, 2017

Here's what I got.

#### Explanation:

Sodium bicarbonate and hydrochloric acid react in a $1 : 1$ mole ratio to produce aqueous sodium chloride, water, and carbon dioxide.

${\text{NaHCO"_ (3(aq)) + "HCl"_ ((aq)) -> "NaCl"_ ((aq)) + "H"_ 2"O"_ ((l)) + "CO}}_{2 \left(g\right)} \uparrow$

This tells you that the reaction will always consume equal numbers of moles of sodium bicarbonate and of hydrochloric acid.

Consequently, you can say that the reactant for which you have the smallest number of moles available will act as the limiting reagent, i.e. it will be completely consumed before all the moles of the second reactant will get the chance to take part in the reaction.

So all you need to do here is to figure out how many moles of each reactant are available.

To find the number of moles of sodium bicarbonate, sue the molar mass of the compound.

2.00 color(red)(cancel(color(black)("g"))) * "1 mole NaHCO"_3/(84.007color(red)(cancel(color(black)("g")))) = "0.0238 moles NaHCO"_3

To find the number of moles of hydrochloric acid, use the molarity and the volume of the solution.

8 color(red)(cancel(color(black)("mL solution"))) * "3.0 moles HCl"/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.024 moles HCl"#

Now, here is where things get a little tricky. Notice that you have one significant figure for the volume of the hydrochloric acid solution, i.e. for $\text{8 mL}$.

This means that you must report the number of moles of hydrochloric acid present in the sample to one significant figure.

$\text{0.024 moles HCl " ~~ " 0.02 moles HCl}$

In this case, you have hydrochloric acid as the limiting reagent because

${\text{0.02 moles HCl " < " 0.0238 moles NaHCO}}_{3}$

However, if you were to take both values rounded to three decimal places, you would have

${\text{0.0238 moles NaHCO"_3 ~~ "0.024 moles NaHCO}}_{3}$

In this case, you'd have equal numbers of moles of both reactants, and thus no limiting reagent.

$\text{0.024 moles NaHCO"_3 = "0.024 moles HCl}$

For example, if you started with $\text{2.00 g}$ of sodium bicarbonate and $\text{8.00 mL}$ of $\text{3.00 M}$ hydrochloric acid solution, you'd round both numbers of moles to three sig figs to get

$\text{0.0238 moles NaHCO"_3 < "0.0240 moles HCl}$

In this scenario, sodium bicarbonate would be the limiting reagent.

This is why significant figures are so important!