# Grams of bicarbonate vs realeased CO2?

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.

• 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.
– Zhe
Sep 19, 2017 at 18:00
• 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, ...
– Karl
Sep 19, 2017 at 20:05
• So, now you apply the stoichiometry from your balanced reaction exactly like Karl says: $\mathrm{mol}\ \ce{NaHCO3} \propto \mathrm{mol}\ \ce{CO2}$.
– Zhe
Sep 19, 2017 at 20:20

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: $\pu{3.05 g}$

• 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. Sep 20, 2017 at 3:07