The fizz produced when an Alka-Seltzer® tablet is dissolved in water is due to the reaction between sodium bicarbonate (NaHCO3) and citric acid (H3C6H5O7): In a certain experiment 1.45g of sodium bicarbonate and 1.45g of citric acid are allowed to react.
How many grams of the excess reactant remain after the limiting reactant is completely consumed?
3NaHCO3+H3C6H5O7-->3CO2+3H2O+Na3C6H5O7
Step-by-step explanation much appreciated with as much detail as possible so I can understand. Thanks!
Use the equation you posted.
Convert 1.45 g each reagent to mols. mols = grams/molar mass
Use the coefficients in the balanced equation to convert mols of each reactant to mols CO2. It is likely that the two numbers will not be the same; obviously one of them is wrong. The correct value in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
Now convert mols of the limiting reagent to mols of the non-limiting reagent (the reagent in excess), change to grams, and subtract from 1.45 g there initially. The difference is the amount that did not react. Post your work if you get stuck.
I was able to figure out that NaHCO3 was the limiting reagent. And then I got that there was .7596 grams of CO2 by converting the 1.45g of NaHCO3. I tried to use 1.45 NaHCO3 to figure it out but, I didn't get the right answer.
My work:
1.45g NaHCO3(1mol NaHCO3/83.998g NaHCO3)(1mol H3C6H5O7/3mol NaHCO3)(192.059g H3C6H5O7/1mol H3C6H5O7)
Can you tell me where I went wrong?
yes.You mixed the steps I gave you.
mol NaHCO3 = 1.45/84(approx) = about 0.173.
mol citric acid = 1.45/(about)192 = about 0.00756.
mols CO2 from NaHCO3 = 0.173 x (3/3) = 0.173
mols CO2 from H3C = 0.00756 x (3/1) = 0.0227
The smaller number is 0.0227 mol; therefore, citric acid is the limiting reagent.
Now convert 0.00756 mols citric acid to mols NaHCO3, convert mols NaHCO3 to grams and subtract from 1.45 there initially.
To find out how many grams of the excess reactant remain after the limiting reactant is completely consumed, we need to follow a step-by-step process.
1. Determine the limiting reactant:
To determine the limiting reactant, we compare the moles of each reactant to the stoichiometric ratio in the balanced equation. The reactant that produces the lowest number of moles of the product is the limiting reactant.
First, let's find the number of moles of each reactant:
Molar mass of sodium bicarbonate (NaHCO3) = 22.99 g/mol (sodium) + 1.01 g/mol (hydrogen) + 12.01 g/mol (carbon) + 48.00 g/mol (oxygen) = 84.01 g/mol
Number of moles of sodium bicarbonate (NaHCO3) = mass / molar mass = 1.45 g / 84.01 g/mol = 0.0172 mol
Molar mass of citric acid (H3C6H5O7) = 1.01 g/mol (hydrogen) + 12.01 g/mol (carbon) + 5 * 1.01 g/mol (hydrogen in alcohol functional groups) + 7 * 16.00 g/mol (oxygen) = 192.13 g/mol
Number of moles of citric acid (H3C6H5O7) = mass / molar mass = 1.45 g / 192.13 g/mol = 0.00754 mol
Next, we can compare the stoichiometric ratio in the balanced equation to determine the limiting reactant:
3NaHCO3 : H3C6H5O7 = 3 : 1
The ratio tells us that 3 moles of sodium bicarbonate react with 1 mole of citric acid. Since we have fewer moles of citric acid (0.00754 mol) compared to sodium bicarbonate (0.0172 mol), citric acid is the limiting reactant.
2. Calculate the amount of product formed:
Now that we know that citric acid is the limiting reactant, we can calculate the amount of product formed based on its mole ratio in the balanced equation.
From the balanced equation, we see that 1 mole of citric acid (H3C6H5O7) reacts to produce 3 moles of carbon dioxide (CO2). Therefore, the number of moles of carbon dioxide produced is 3 * 0.00754 mol = 0.0226 mol.
3. Convert moles of reactant to grams of remaining excess reactant:
Now we need to calculate the moles of sodium bicarbonate (NaHCO3) that would react completely with the limiting reactant (0.00754 mol of citric acid).
From the stoichiometric ratio, we know that 3 moles of sodium bicarbonate react with 1 mole of citric acid. Therefore, the amount of sodium bicarbonate that would react with 0.00754 mol of citric acid is 3 * 0.00754 mol = 0.0226 mol.
To calculate the mass of the excess reactant (sodium bicarbonate) that remains, we can subtract the moles of sodium bicarbonate used (0.0226 mol) from the total moles of sodium bicarbonate given (0.0172 mol).
Moles of excess sodium bicarbonate = total moles of sodium bicarbonate - moles of sodium bicarbonate used = 0.0172 mol - 0.0226 mol = -0.0054 mol
The negative value indicates that the excess reactant is completely consumed during the reaction. Therefore, no grams of excess sodium bicarbonate remain after the limiting reactant is completely consumed.
In summary, after the limiting reactant (citric acid) is completely consumed, there are no grams of excess reactant (sodium bicarbonate) remaining.