If you start using 10.0 grams of NaHCO3 and excess of acetic acid, how many moles of CO2 can be formed? How many grams of CO2 can be formed?

And 10.0 grams of sodium bicarbonate (molar mass = 84 g/mol) are reacting in a 1:1 mole ratio with the yielded carbon dioxide, those moles and grams turn out:

[tex]n_{CO_2}=10.0gNaHCO_3*\frac{1molNaHCO_3}{84gNaHCO_3} *\frac{1molCO_2}{1molNaHCO_3} =0.119molCO_2[/tex]

And the mass by considering the molar mass of carbon dioxide to be 44 g/mol:

[tex]m_{CO_2}=0.119molCO_2*\frac{44gCO_2}{1molCO_2} \\\\m_{CO_2}=5.24gCO_2[/tex]

Best regards.

dsdrajlin dsdrajlin · Answer 2 · 2021-10-22T14:22:26+02:00

If you start using 10.0 grams of sodium bicarbonate and excess of acetic acid, you can obtain 0.119 moles of CO₂ (5.24 g).

Let's consider the following neutralization reaction.

NaHCO₃ + CH₃COOH ⇒ CH₃COONa + H₂O + CO₂

First, we will convert 10.0 g of NaHCO₃ to moles using its molar mass (84.01 g/mol).

[tex]10.0 g \times \frac{1mol}{84.01g} = 0.119mol[/tex]

The molar ratio of NaHCO₃ to CO₂ is 1:1. The moles of CO₂ formed from 0.119 moles of NaHCO₃ are:

[tex]0.119 mol NaHCO_3 \times \frac{1molCO_2}{1molNaHCO_3} = 0.119 moles CO_2[/tex]

Finally, we will convert 0.119 moles of CO₂ to grams using its molar mass (44.01 g/mol).

[tex]0.119 mol \times \frac{44.01g}{mol} = 5.24 g[/tex]

If you start using 10.0 grams of sodium bicarbonate and excess of acetic acid, you can obtain 0.119 moles of CO₂ (5.24 g).

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