Question

I have having trouble with data table 4 and i would like someone to double check my work on the other tables.

NaHCO3(s) + HC2H3O2(aq) → C2H3O2Na(aq) +H2CO3(aq) H2CO3(aq)→ H2O(l) +CO2(g)

Note: 5 mL of 5% vinegar contains 0.25 mL of HC₂H₃O₂ with a density of 1.0 g/mL which equals 0.25 g HC₂H₃O₂.

Data Table 4

                          Limiting Reactant            Theoretical yield of CO₂(moles)    Theoretical yield of CO₂(g)

Reaction 1        NaHCO₂

Reaction 2        NaHCO₂

Reaction 3       HC₂H₃O₂

Reaction 4        HC₂H₃O₂

_{Data Table 2}

                          Volume of 5%            Mass of HC₂H₃O₂                                            Moles of HC₂H₃O₂

                          Vinegar (mL)

Reaction 1        5.0                                        0.10                                                      0.004

Reaction 2        5.0                                        0.20                                                      0.004

Reaction 3       5.0                                        0.35                                                     0.004

Reaction 4        5.0                                        0.50                                                     0.004

Data Table 1

  Mass of NaHCO₃(g)                        Moles of NaHCO₃

Reaction 1     0.10                                                      0.001

Reaction 2     0.20                                                      0.002

Reaction 3    0.35                                                     0.004

Reaction 4    0.50                                                     0.005

Answer 1

The reaction is

NaHCO₃ + HC₂H₃O₂ NaC₂H₃O₂ + H₂O + CO₂

Thus one mole of NaHCO₃ reacts with one mole of HC₂H₃O₂ gives one mole of CO₂

5 mL of 5% vinegar contains 0.25 g of HC₂H₃O₂ = 0.25 / 60 = 0.00417 mole

[ here 60 is the molar mass of HC₂H₃O₂ ]

Since in each reaction, the same quantity of 5 mL vinegar is taken, the moles of HC₂H₃O₂ is the same for the four reactions.

Table 2
Reaction	mass of	moles of		mass of	moles of
	HC2H3O2	HC2H3O2		NaHCO3	NaHCO3
1	0.25 g	0.00417		0.10	0.00119
2	0.25 g	0.00417		0.20	0.00238
3	0.25 g	0.00417		0.35	0.00417
4	0.25 g	0.00417		0.50	0.00595

the mass of NaHCO₃ for the first reaction is 0.10 g = 0.10 / 84 = 0.00119 moles

[ here 84 is the molar mass of NaHCO₃]

similarly for reaction 2,    0.2 / 84 = 0.00238 moles

                 reaction 3    0.35/84 = 0.00417 moles

                reaction 4             0.50 / 84 = 0.00595 moles

Comparing the moles of NaHCO₃ and moles of HC₂H₃O₂ for all the four reactions,

in reaction 1 and in reaction 2, the reactant NaHCO₃ is present less than that of HC₂H₃O₂ and hence NaCO₃ is the limiting reactant.

in reaction 3, both are equal and hence both are limiting reactants.and there is no excess reactant.

in reaction 4, obviously HC₂H₃O₂ is the limiting reactant..

Table 4
Reaction	Limiting	Theoretical yield
	Reactant	CO2 - moles	CO2 - g
1	NaHCO3	0.00119	0.052 g
2	NaHCO3	0.00238	0.105 g
3	--	0.00417	0.183 g
4	HC2H3O2	0.00417	0.183 g

The moles of CO₂ formed is the moles of limiting reactant. and the mass of CO₂ is calculated by multiplying the moles by 44, the molar mass of CO₂

Thus for reaction 1, moles of CO2 = 0.00119 mole = 0.00119 mol x 44 g / mol = 0.052 g

similarly for reaction 2              0.00238 x 44 = 0.105 g

            for reaction 3              0.00417 x 44 = 0.183 g

            for reaction 4              0.00417 x 44 = 0.183 g