Using the formula:

\[ \text{NaHCO}_3 (s) + \text{HC}_2\text{H}_3\text{O}_2 (aq) \rightarrow \text{NaC}_2\text{H}_3\text{O}_2 (aq) + \text{H}_2\text{O} (l) + \text{CO}_2 (g) \]

If I used 0.100 L of 0.83 M \(\text{HC}_2\text{H}_3\text{O}_2 (aq)\) (0.83 mol per 1 L of solution), how much carbon dioxide would be produced?

Moles=volume concentration =0.1 .83 =.083 Moles of HC2H3O2 Mole ratio between HC2H3O2 and CO2 is 1:1 This means .083 Moles of CO2
Mass =Moles Rfm of CO2 =.083 (12+16+16) =3.7grams

Answered by Jerboa | 2024-06-10

By utilizing stoichiometry , we can determine that the given volume and molarity of acetic acid (HC2H3O2) would produce approximately 3.65 grams of carbon dioxide (CO2). The calculation involves determining the moles of HC2H3O2 used, which equals the moles of CO2 produced, and converting that to grams using the molar mass of CO2. ;

Answered by adarshjanise46 | 2024-06-18

By using stoichiometry, we can determine that 0.100 L of 0.83 M acetic acid will produce approximately 3.65 grams of carbon dioxide. This was calculated using the moles of acetic acid, which directly corresponds to the moles of carbon dioxide produced based on the balanced equation. Thus, the total amount of CO₂ produced is 3.65 g.
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Answered by adarshjanise46 | 2024-10-10