Question

The lab experiment instructs you to react 0.21g of NaHCO3 with excess CH3COOH. How CO2(g) in mL would this reaction generate if all the sodium bicarbonate reacts fully? (assume the room temp is 25)

Answer 1

When 0.21g of NaHCO3 reacts fully with excess CH3COOH, it will produce 112 mL of CO2(g).

Step-by-step explanation:

To determine the volume of CO2(g) generated when 0.21g of NaHCO3 reacts fully with excess CH3COOH, we need to consider the balanced chemical equation:

2 NaHCO3(aq) + 2 CH3COOH(aq) → 2 CH3COONa(aq) + 2 H2O(ℓ) + 2 CO2(g)

From the equation, we can see that for every 2 moles of NaHCO3, 2 moles of CO2 are produced. The molar mass of NaHCO3 is 84 g/mol, so 0.21 g is equivalent to 0.21 g / 84 g/mol = 0.0025 mol.

Therefore, if all the NaHCO3 reacts fully, it will produce 0.0025 mol × 2 mol CO2/mol NaHCO3 = 0.005 mol CO2. Since 1 mol of any ideal gas occupies 22.4 L at STP (standard temperature and pressure), the volume of CO2 generated would be 0.005 mol × 22.4 L/mol = 0.112 L or 112 mL.