A baker uses sodium hydrogen carbonate (baking soda) as the...

A baker uses sodium hydrogen carbonate (baking soda) as the leavening agent in a banana-nut quickbread. The baking soda decomposes in either of two possible reactions: $$ \begin{array}{l}{\text { (1) } 2 \mathrm{NaHCO}_{3}(s) \longrightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g)} \\ {\text { (2) } \mathrm{NaHCO}_{3}(s)+\mathrm{H}^{+}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g)+\mathrm{Na}^{+}(a q)}\end{array} $$ Calculate the volume (in mL) of $\mathrm{CO}_{2}$ that forms at $200 .^{\circ} \mathrm{C}$ and 0.975 atm per gram of $\mathrm{NaHCO}_{3}$ by each of the reaction processes.

A baker uses sodium hydrogen carbonate (baking soda) as the leavening agent in a banana-nut quickbread. The baking soda decomposes in either of two possible reactions:
$$
\begin{array}{l}{\text { (1) } 2 \mathrm{NaHCO}_{3}(s) \longrightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g)} \\ {\text { (2) } \mathrm{NaHCO}_{3}(s)+\mathrm{H}^{+}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g)+\mathrm{Na}^{+}(a q)}\end{array}
$$
Calculate the volume (in mL) of $\mathrm{CO}_{2}$ that forms at $200 .^{\circ} \mathrm{C}$ and 0.975 atm per gram of $\mathrm{NaHCO}_{3}$ by each of the reaction processes.

Key Concepts

Stoichiometry

Stoichiometry involves using balanced chemical equations to determine the quantitative relationships between reactants and products. In the context of gas evolution reactions, knowing the mole ratios from the balanced equations is essential for calculating how many moles—and thus what volume—of CO2 is produced per given amount of sodium hydrogen carbonate.

Ideal Gas Law

The Ideal Gas Law (PV = nRT) is a fundamental principle that relates the pressure, volume, temperature, and number of moles of a gas. It is used to calculate the volume of gas generated under specific conditions by substituting the known values of temperature, pressure, and moles of CO2 derived from the stoichiometry of the reaction.

Decomposition Reactions

Decomposition reactions involve the breakdown of a single compound into multiple products. In chemical processes like the decomposition of sodium hydrogen carbonate, understanding the mechanism of decomposition is critical for determining the type and amount of gases evolved, which in turn affects the calculations for gas volumes.

Reaction Mechanisms

Reaction mechanisms describe the step-by-step process by which reactants transform into products. In cases where a compound can decompose via different pathways—as seen with sodium hydrogen carbonate—the mechanism determines the specific stoichiometry of the reaction and consequently the amount of CO2 produced. This insight is crucial when comparing different reaction pathways.

Transcript

00:01 So for this question, they are wanting us to find the volume of co2 per gram of sodium bicarbonate for both reactions 1 and 2.

00:11 So these are both balanced out equations.

00:14 So we're going to be using our equation pv is equal to nrt to help us solve this problem.

00:22 For this, we're looking for our volume of co2 in milliliters.

00:27 It's the volume of co2 and milliliters per gram of our sodium bicarbonate.

00:36 They tell us that the pressure for this reaction is 0 .975 atmospheres.

00:45 And they also tell us that the temperature for this reaction is 200 degrees c.

00:50 So i'm just going to add 273 kelvin to turn this into kelvin.

00:55 So, 473 kelvin.

01:00 And i'm going to be using the r value that's equal to 0 .08 to 1 liters atmosphere per mole kelvin.

01:12 And we're also going to be using the molar mass.

01:14 I'm just going to abbreviate that mm of the sodium bicarbonate.

01:20 And that's going to equal 84 .007 grams per.

01:27 Mole.

01:29 All right, so to figure out the volume of co2 per gram sodium bicarb for reaction one, we see that for every two moles of our sodium bicarbonate, we get one mole of our co2.

01:55 So to get the moles of co2 per gram.

01:59 We're going to take one mole of our sodium bicarb and we're going to divide that by its mass or molar mass so 84 .007 grams and then i'm going to times that by the ratio so one mole co2 for every two moles of our bicarbonate.

02:26 So when we calculate this out, this is going to be 0 .00592 moles of co2 per gram of our sodium bicarbonate.

02:49 All right, so to find the volume of our co2 per gram, we're just going to take the moles that we just found.

02:59 So 0 .005952 moles co2 per gram...

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