1
$\begingroup$

So, it's been a good two years since I've done chemistry and I have my first university lab in a few days and I'm sort of freaking out because I can't even remember how to do stoichiometry properly. This a question off of my pre-lab review and I'm having troubles remembering what to do. I believe this is a limiting reagent question.

In Experiment 1, you will place and AlkaSeltzer tablet in each of the dilute $\ce{HCl}$ solutions you prepared. The reaction between $\ce{HCl}$ and $\ce{NaHCO3}$ should take place quickly and release gaseous $\ce{CO2}$. You will then use the mass lost for each flask to determine the amount of $\ce{NaHCO3}$ that reacted.

A student performs this experiment and determines that $\pu{1.60 g}$ of $\ce{CO2}$ were lost. How many grams of $\ce{NaHCO3}$ would have reacted to produce this amount of $\ce{CO2}$?

Firstly, I wrote out part of the equation but I'm also beginning to think it's incorrect too:

$$\ce{NaHCO3(aq) + HCl(aq) <=> CO2(g) + NaCl + H2O(l)}$$

The initial amount for $\ce{NaHCO3}$, I know, is unknown and the final amount for $\ce{CO2}$ is $\pu{1.60 g}$. Other than that though I'm at a loss for what to do next. Please help if you can, thank you!

Edit (see answer): To calculate the amount of moles in $\ce{CO2(g)}$ I used the formula $m/M = n$, which gave me 0.36355 moles for carbon dioxide. Now I use 0.36355 moles for $\ce{NaHCO3(aq)}$ since it's a 1:1 ratio.

$\endgroup$
  • $\begingroup$ For starters, carbon dioxide is $\ce{CO2}$ not $\ce{CO3}$. The products of the reaction are sodium chloride (which you have), carbon dioxide (not quite right in your equation), and water (we don't make hydrogen). I would start there, and try to fix your balanced equation. $\endgroup$ – Zhe Sep 19 '17 at 18:00
  • $\begingroup$ x grams of CO2 are so-many moles of CO2, the same amount of moles of carbonate are on the other side of the equation, ... $\endgroup$ – Karl Sep 19 '17 at 20:05
  • $\begingroup$ So, now you apply the stoichiometry from your balanced reaction exactly like Karl says: $\mathrm{mol}\ \ce{NaHCO3} \propto \mathrm{mol}\ \ce{CO2}$. $\endgroup$ – Zhe Sep 19 '17 at 20:20
5
$\begingroup$

Thank you for all the help, I've figured out the answer:

$$\ce{NaHCO3(aq) + HCl(aq) -> NaCl(aq) + CO2(g) + H2O(l)}$$

Since we're given $\pu{1.60 g}$ of $\ce{CO2(g)}$ we can use the mass to calculate the number of moles with the equation:

$$n = \frac{m}{M} = \frac{\pu{1.60 g}}{\pu{44.01 g mol-1}} = \pu{0.036355 mol}$$

Since it's a 1 : 1 ratio between $\ce{NaHCO3(aq)}$ and $\ce{CO2(g)}$, we can directly solve for the mass of $\ce{NaHCO3(aq)}$ using the same formula:

$$n = \frac{m}{M} $$

$$m = n \times M = \pu{0.036355 mol} \times \pu{83.42 g mol-1} = \pu{3.0509116 g}$$

Round to two decimal places: $\pu{3.05 g}$

$\endgroup$
  • 1
    $\begingroup$ Great work! Keep in mind you can accept your own answer. As an advice, it's generally better to round up to significant digits at once. $\endgroup$ – andselisk Sep 20 '17 at 3:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.