Question

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?

Answer 1

Answer:

$n_{CO_2}=0.119molCO_2$

$m_{CO_2}=5.24gCO_2$

Explanation:

Hello.

In this case, since the chemical reaction:

$NaHCO_3+CH_3COOH\rightarrow CH_3COONa+H_2O+CO_2$

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:

$n_{CO_2}=10.0gNaHCO_3*\frac{1molNaHCO_3}{84gNaHCO_3} *\frac{1molCO_2}{1molNaHCO_3} =0.119molCO_2$

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

$m_{CO_2}=0.119molCO_2*\frac{44gCO_2}{1molCO_2} \\\\m_{CO_2}=5.24gCO_2$

Best regards.