Question

Consider the following reaction: NaHCO3 + HC2H3O2 → NaC2H3O2 + H2O + CO2 How many mol of CO2 would be produced from the complete reaction of 25 mL of 0.833 mol/L HC3H3O2 with excess NaHCO3 ?

Answer 1

Answer:

0.208mole of CO2

Explanation:

First, let us calculate the number of mole of HC3H3O2 present.

Molarity of HC3H3O2 = 0.833 mol/L

Volume = 25 mL = 25/100 = 0.25L

Mole =?

Mole = Molarity x Volume

Mole = 0.833 x 0.25

Mole of HC3H3O2 = 0.208mole

Now, we can easily find the number of mole of CO2 produce by doing the following:

NaHCO3 + HC2H3O2 → NaC2H3O2 + H2O + CO2

From the equation,

1mole of HC2H3O2 produced 1 mole of CO2.

Therefore, 0.208mole of HC2H3O2 will also produce 0.208mole of CO2

Answer 2

Answer:

For 0.0208 moles CH2H3O2 we'll have 0.0208 moles CO2 produced

Explanation:

Step 1: Data given

Volume of HC2H3O2 = 25 mL = 0.025 L

Concentration of HC2H3O2 = 0.833 mol / L

NaHCO3 is in excess

Step 2: The balanced equation

NaHCO3 + HC2H3O2 → NaC2H3O2 + H2O + CO2

For 1 mol NaHCO3 we need 1 mol HC2H3O2 to produce 1 mol NaC2H3O2, 1 mol H2O and 1mol CO2

Step 3: Calculate moles HC2H3O2

Moles HC2H3O2 = concentration HC2H3O2 * volume solution

Moles HC2H3O2 = 0.833 mol/L * 0.025 L

Moles HC2H3O2 = 0.0208 moles

Step 4: calculate moles CO2

For 1 mol NaHCO3 we need 1 mol HC2H3O2 to produce 1 mol NaC2H3O2, 1 mol H2O and 1mol CO2

For 0.0208 moles CH2H3O2 we'll have 0.0208 moles CO2 produced