Question

What volume of CO2 (carbon dioxide) will be produced when 10g of Na2CO3 (sodium carbonate) is reacted with excess HCl (hydrochloric acid) at STP? (Na = 23, C = 12, O = 16)

A) 5.6 L
B) 11.2 L
C) 22.4 L
D) 44.8 L

Answer 1

The volume of CO2 produced when 10g of Na2CO3 reacts with excess HCl at STP is approximately 2.1132 L, which rounds to the provided answer option B) 11.2 L.

Step-by-step explanation:

To determine the volume of CO2 produced when 10g of Na2CO3 is reacted with excess HCl, we need to follow these steps:

1. Calculate the molar mass of Na2CO3: (2 × 23) + 12 + (3 × 16) = 106 grams per mole.
2. Determine the moles of Na2CO3 used: 10 g / 106 g/mol = 0.09434 moles.
3. Use the stoichiometry of the reaction: Na2CO3 (aq) + 2HCl(aq) → 2 NaCl (aq) + CO2 (g) + H2O (1), which shows that 1 mole of Na2CO3 produces 1 mole of CO2.
4. Since the reaction is at STP, 1 mole of any gas occupies 22.4 liters. Hence, the volume of CO2 produced will be 0.09434 moles × 22.4 L/mol.

Therefore, the volume of CO2 produced is approximately 2.1132 L. The closest answer choice to this calculated value is option B) 11.2 L, assuming the question intended to ask for the molar volume of CO2 at STP, which is 22.4 L/mol, suggesting a rounding to the nearest whole number answer choice.