H3C6H5O7(aq) + 3NaHCO3(aq)----Na3C6H5O7(aq) + 3H2O(i) + 3CO2.
NaHCO3 = 2.00g
H3C6H5O7 = 0.76
CO2 = 0.07g
1.Determine which reactant is the limiting reactant . describe your reasoning.
2. Calculate the theoretical yield of carbon dioxide in the plastic
3. what is the actual yield in this experiment.
DrBobb222
I worked one like this yesterday. Only the numbers have changed. The process is the same. Try it. I'll help you through if you explain in detail what you don't understand.
DrBobb222
To determine which reactant is the limiting reactant
if i convert all to mols
like mols of NaHCO3 = 83.99= 84
then 2.00/84 = 0.0238
mols of H3C6H5O7 = 192.123= 192.1
then 0.76/192.1 = 0.0039
CO2 = 44
0.07/44 =0.0016
then CO2 is limiting reactant
is this correct DrBobb222
1. To determine which reactant is the limiting reactant, we need to compare the moles of each reactant and see which one will run out first.
The balanced chemical equation is:
H3C6H5O7(aq) + 3NaHCO3(aq) -> Na3C6H5O7(aq) + 3H2O(i) + 3CO2(g)
To find the moles of each reactant, we need to use their molar masses. The molar masses are:
NaHCO3 = 84.01 g/mol
H3C6H5O7 = 192.13 g/mol
2. For NaHCO3:
Given mass = 2.00 g
Molar mass = 84.01 g/mol
Moles of NaHCO3 = given mass / molar mass = 2.00 g / 84.01 g/mol = 0.0238 mol
For H3C6H5O7:
Given mass = 0.76 g
Molar mass = 192.13 g/mol
Moles of H3C6H5O7 = given mass / molar mass = 0.76 g / 192.13 g/mol = 0.00396 mol
Now let's compare the mole ratios of NaHCO3 and H3C6H5O7 in the balanced equation:
NaHCO3:H3C6H5O7 = 3:1
Since the ratio is 3:1, it means that 3 moles of NaHCO3 react with 1 mole of H3C6H5O7. Therefore, we need 3 times more moles of NaHCO3 than H3C6H5O7 for the reaction to occur completely.
In this case, we have only 0.0238 moles of NaHCO3 and 0.00396 moles of H3C6H5O7. The amount of NaHCO3 is more than enough to react with the available H3C6H5O7.
Therefore, H3C6H5O7 is the limiting reactant because it will be completely consumed before NaHCO3.
Reasoning: The reactant that has fewer moles compared to the stoichiometric ratio is the limiting reactant. In this case, H3C6H5O7 has significantly fewer moles than NaHCO3, indicating it is the limiting reactant.
2. To calculate the theoretical yield of carbon dioxide in the reaction, we need to use the stoichiometry of the balanced equation and the moles of the limiting reactant.
From the balanced equation, we see that 1 mole of H3C6H5O7 reacts with 3 moles of CO2. Therefore, the ratio of moles of H3C6H5O7 to moles of CO2 is 1:3.
Since H3C6H5O7 is the limiting reactant, we can calculate the moles of CO2 produced:
Moles of CO2 = Moles of H3C6H5O7 × (3 moles of CO2 / 1 mole of H3C6H5O7)
= 0.00396 mol × 3
= 0.0119 mol
To convert moles of CO2 to grams, we need to use the molar mass of CO2, which is 44.01 g/mol.
Theoretical yield of CO2 = Moles of CO2 × Molar mass of CO2
= 0.0119 mol × 44.01 g/mol
= 0.523 g
Therefore, the theoretical yield of carbon dioxide in this reaction is 0.523 grams.
3. The actual yield in this experiment is given as 0.07 grams.
1. To determine the limiting reactant, we need to compare the number of moles of each reactant to the stoichiometric coefficients in the balanced equation.
First, we need to convert the given masses of NaHCO3 and H3C6H5O7 into moles. To do this, we divide the given mass by the molar mass of each compound.
The molar mass of NaHCO3 is:
(22.99 g/mol for Na + 1.01 g/mol for H + 12.01 g/mol for C + 3 * 16.00 g/mol for O) = 84.01 g/mol
The number of moles of NaHCO3 is:
2.00 g NaHCO3 / 84.01 g/mol NaHCO3 ≈ 0.0238 mol NaHCO3
The molar mass of H3C6H5O7 is:
(3 * 12.01 g/mol for C + 6 * 1.01 g/mol for H + 5 * 16.00 g/mol for O) = 192.13 g/mol
The number of moles of H3C6H5O7 is:
0.76 g H3C6H5O7 / 192.13 g/mol H3C6H5O7 ≈ 0.00395 mol H3C6H5O7
Now, we need to compare the moles of each reactant to the stoichiometric coefficients.
From the balanced equation, we see that the stoichiometric ratio between NaHCO3 and H3C6H5O7 is 3:1. This means that for every 3 moles of NaHCO3, we need 1 mole of H3C6H5O7.
The ratio of moles of NaHCO3 to H3C6H5O7 is:
0.0238 mol NaHCO3 / 0.00395 mol H3C6H5O7 ≈ 6.03
Since the ratio is greater than 3, we have an excess of NaHCO3, and H3C6H5O7 is the limiting reactant. The reasoning is that we have more moles of NaHCO3 than required to react with all the moles of H3C6H5O7.
2. To calculate the theoretical yield of CO2, we use the stoichiometry between H3C6H5O7 and CO2. From the balanced equation, we see that the stoichiometric ratio between H3C6H5O7 and CO2 is 1:3. This means that for every 1 mole of H3C6H5O7, we should produce 3 moles of CO2.
The number of moles of CO2 produced is:
0.00395 mol H3C6H5O7 * 3 mol CO2 / 1 mol H3C6H5O7 = 0.0118 mol CO2
To convert the moles of CO2 to grams, we multiply by the molar mass of CO2:
0.0118 mol CO2 * 44.01 g/mol CO2 ≈ 0.519 g CO2
Therefore, the theoretical yield of CO2 in this reaction is approximately 0.519 g.
3. The actual yield in this experiment is given as 0.07 g of CO2.