If sodium hydrogen carbonate is added to a hydrochloric acid solution, it will produce gaseous carbon dioxide according to the following reaction:

NaHCO3(s) + HCl(aq) → NaCl(aq) + H2O(l) + CO2(g)

How many grams of NaHCO3 must be added to an excess of HCl(aq) to produce 50.0 mL of CO2 at 25 °C and 0.995 atm?

Question

If sodium hydrogen carbonate is added to a hydrochloric acid solution, it will produce gaseous carbon dioxide according to the following reaction:

NaHCO3(s) + HCl(aq) → NaCl(aq) + H2O(l) + CO2(g)

How many grams of NaHCO3 must be added to an excess of HCl(aq) to produce 50.0 mL of CO2 at 25 °C and 0.995 atm?

BlackTom AI · Accepted Answer

To tackle this problem, start by identifying the gas produced and the conditions given. The reaction shows that 1 mole of NaHCO3 yields 1 mole of CO2, so the amount of NaHCO3 needed will equal the moles of CO2 required, based on the stoichiometry.

First, compute the moles of CO2 corresponding to 50.0 mL at 25 °C and 0.995 atm using the ideal gas law: n = PV/RT. Here P = 0.995 atm, V = 0.0500 L, R = 0.082057 L·atm/(mol·K), and T = 298 K (25 °C).

Calculate the numerator: P × V = 0.995 × 0.0500 = 0.04975. Calculate the denominator: R × T = 0.082057 × 298 ≈ 24.453. Therefore n(CO2) ≈ 0.04975 / 24.453 ≈ 2.04 × 10^-3 mol.

Since the reaction is 1:1, n(NaHCO3) = n(CO2) ≈ 2.04 × 10^-3 mol.

Next, convert moles of NaHCO3 to grams using its molar mass. NaHCO3 has a molar mass of about 84.01 g/mol. Mass = moles × molar mass ≈ (2.04 × 10^-3) × 84.01 g/mol ≈ 0.171 g.

Now evaluate the provided options:
- 171 g: This is far too large by about three orders of magnitude. The calculation shows only millimoles are needed, not hundreds of grams.
- 2.04 g: This is roughly 1000 times larger than the correct value in grams, reflecting a misplacement of decimal or miscalculation of moles-to-grams.
- 0.00203 g: This corresponds to the moles (2.03 × 10^-3 mol) treated as grams, which ignores the molar mass conversion entirely and is far too small.
- 0.170 g: This aligns with the calculated mass (about 0.171 g) when converting 2.04 × 10^-3 mol to grams using 84.01 g/mol; it matches the stoichiometry and gas-law calculation.

Why the other options fail can be traced to common mistakes: misinterpreting the units (confusing moles with grams), forgetting the molar mass, or misapplying the ideal gas law constants leading to incorrect n(CO2).

In summary, accurately applying PV = nRT to find moles of CO2, then using a 1:1 stoichiometric ratio to NaHCO3, followed by converting moles to grams with the molar mass yields approximately 0.171 g of NaHCO3 needed.

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