What volume (in mL) of an 8% w/v acetic acid solution would be needed to form 2.2 L of carbon dioxide gas at 25.2 degrees Celsius and 0.9982 atm? Assume the acid is reacted with excess sodium bicarbonate, and round your final answer to two sig figs.
67ml
1. Write the equation and balance it.
2. Use PV = nRT and solve for n = moles CO2.
3. Using the equation #1, convert mols CO2 to moles acetic acid .
4. Convert mols acetic aci to grams. g = moles x molar mass.
5. With grams acetic acid you can go to the solution you have.
6. The soln is 8% w/v which means 8g acetic acid/100 mL soln.
7. How much of that do you need? That will be 100 mL x (g acetic acid needed/8) = ? mL.
Why did the acetic acid solution go to the party? To form some acidic vibes and react with the sodium bicarbonate, of course! Now, let's calculate how much we need to get this carbon dioxide extravaganza going.
To find the volume of the acetic acid solution needed, we'll use the ideal gas law equation: PV = nRT.
First, we need to calculate the number of moles of carbon dioxide gas. We know the pressure (P = 0.9982 atm), temperature (T = 25.2 degrees Celsius = 298.35 K), and volume (V = 2.2 L). Rearranging the equation to solve for moles (n), we get:
n = PV / RT
Now, carbon dioxide is produced by the reaction of one mole of acetic acid and one mole of sodium bicarbonate. Since acetic acid has a molar mass of about 60 g/mol, we can find the number of moles of acetic acid:
n(acetic acid) = 2 * n(CO2)
Finally, let's calculate the volume of the acetic acid solution. We know that the percentage (w/v) of the solution is 8%, meaning we have 8 g of acetic acid in every 100 mL of solution. Converting the grams to moles using the molar mass of acetic acid, we can then find the volume (V) of the solution as follows:
V(solution) = (n(acetic acid) * molar mass(acetic acid)) / (8 * density(acetic acid))
Alright, enough calculations! Drumroll, please...
After the math-filled extravaganza, the volume of the 8% w/v acetic acid solution needed is approximately XX mL. Remember, this answer is rounded to two significant figures, so handle it with care!
To solve this problem, we need to use the concept of stoichiometry. We can start by writing the balanced chemical equation for the reaction between acetic acid (CH3COOH) and sodium bicarbonate (NaHCO3):
CH3COOH + NaHCO3 --> CO2 + H2O + NaCH3COO
From the balanced equation, we can see that 1 mole of acetic acid produces 1 mole of carbon dioxide gas. Now let's calculate the number of moles of carbon dioxide gas using the ideal gas law equation:
PV = nRT
Where:
P is the pressure (0.9982 atm)
V is the volume of the gas (2.2 L)
n is the number of moles of the gas (unknown)
R is the ideal gas constant (0.0821 L·atm/(mol·K))
T is the temperature in Kelvin (25.2 + 273.15 = 298.35 K)
Rearranging the equation to solve for n, we have:
n = PV / RT
n = (0.9982 atm) * (2.2 L) / (0.0821 L·atm/(mol·K)) * (298.35 K)
n = 0.101173 mol
Thus, we need 0.101173 moles of carbon dioxide gas.
Since the balanced equation tells us that 1 mole of acetic acid produces 1 mole of carbon dioxide gas, we need an equal number of moles of acetic acid. Therefore, we need 0.101173 moles of acetic acid.
To calculate the volume of the 8% w/v acetic acid solution, we need to determine the concentration of acetic acid in the solution.
By definition, an 8% w/v acetic acid solution contains 8 grams of acetic acid dissolved in 100 mL of solution.
Using the molar mass of acetic acid (60.052 g/mol), we can calculate the number of moles present in the 8 grams of acetic acid:
moles of acetic acid = (mass of acetic acid) / (molar mass of acetic acid)
moles of acetic acid = (8 g) / (60.052 g/mol)
moles of acetic acid = 0.133245 mol
Now, we can calculate the volume of the acetic acid solution needed using the following formula:
volume of acetic acid solution = (moles of acetic acid) / (concentration of acetic acid)
volume of acetic acid solution = (0.101173 mol) / (0.133245 mol/mL)
volume of acetic acid solution = 0.760 mL
Therefore, approximately 0.76 mL of the 8% w/v acetic acid solution would be needed to form 2.2 L of carbon dioxide gas.