[johnknee]

In experiment 1, you study a neutralization reaction involving an acid and a base.

Oxalosuccinic acid (H3C6H3O7) is a tricarboxylic acid.

It reacts with NaHCO3 (sodium bicarbonate or "baking soda") to yield carbon dioxide, water and sodium oxalosuccinate (Na3C6H3O7).

Balance the following reaction:

H3C6H3O7(aq) + 3NaHCO3(s) = 3CO2(g) + 3H2O(l) + Na3C6H3O7(aq)

A student measures 11.41 g of baking soda and adds this to 24. mL of 3.0 M oxalosuccinic acid (in an alcoholic solution).

Identify the limiting reagent and calculate the number of moles of CO2 produced.

The limiting reagent is:
sodium bicarbonate.

1.358×10-1 mol of CO2 are produced.

QUESTION: Hypothetically, if the temperature of the solution goes from 25.0°C to 8.0°C the reaction is considered

Calculate the temperature change per mole of H+ which reacted.

ΔT/(mol H+ reacted) = K/mol

*11.41 g NaHCO3/84.007 g/mol = 0.1358 mol NaHCO3.
3 M * 0.24 L = 0.72 mol H3C6H3O7

I determined that baking soda was the limiting reagent through some calculations and verifIcations.

For the question I am stuck on, I know that Delta T is -17 from 8-25.
I am not quite sure how to get the moles of H+.

Please any help is appreciated.

[Dan]

You must show you have attempted the question, this is a Forum Rule.

[Burner]

H3C6H3O7(aq) + 3NaHCO3(s) = 3CO2(g) + 3H2O(l) + Na3C6H3O7(aq)

A student measures 11.41 g of baking soda and adds this to 24. mL of 3.0 M oxalosuccinic acid (in an alcoholic solution).

The limiting reagent is:
sodium bicarbonate.

1.358×10-1 mol of CO2 are produced.

[...]

For the question I am stuck on, I know that Delta T is -17 from 8-25.
I am not quite sure how to get the moles of H+.

Since you found that sodium bicarbonate is the limiting reagent, you need to use the number of moles of sodium bicarbonate and the mole ratio from the equation to find out the number of moles of H₃C₆H₃O₇ reacted first, then go on to calculate the number of moles of H⁺.

[johnknee]

Yeah I believe I found the moles of  H3C6H3O7 reacted through the mole bridge to be 0.045 mol.

>> 0.1358 mol NaHCO3 * (1mol  H3C6H3O7/3mol NaHCO3) = 0.045 mol  H3C6H3O7

>> also 0.72 mol  H3C6H3O7 * (3mol NaHCO3/1mol  H3C6H3O7) = 2.16 mol NaHCO3

 Would I simply multiply 0.045 by 3 to find moles of H+?

[Burner]

Yeah I believe I found the moles of  H3C6H3O7 reacted through the mole bridge to be 0.045 mol.

>> 0.1358 mol NaHCO3 * (1mol  H3C6H3O7/3mol NaHCO3) = 0.045 mol  H3C6H3O7

>> also 0.72 mol  H3C6H3O7 * (3mol NaHCO3/1mol  H3C6H3O7) = 2.16 mol NaHCO3

 Would I simply multiply 0.045 by 3 to find moles of H+?

Yes