If you wanted to produce 0.25 L of CO2 gas (at STP), how many grams of sodium bicarbonate would I need to use? Acetic acid is in excess. (IT IS ASSUMED YOUR ANSWER IS IN g NaHCO3 so you do not need to include that in your answer.)
Acetic acid is HC2H3O2 OR CH3COOH or HAc.
NaHCO3 + HAc ==> NaAc + H2O + CO2
mols CO2 gas in 0.25L is 0.25L x (1 mol/22.5L) = approx 0.0112
Using the coefficients in the balanced equation, convert mols CO2 to mols NaHCO3. That's 0.0112 mols CO2 x (1 mol NaHCO3/1 mol CO2) = 0.0112 x 1/1 = 0.0112 mols NaHCO3.
Convert mols NaHCO3 to grams = mols x molar mass.
To determine the amount of sodium bicarbonate (NaHCO3) needed to produce 0.25 L of carbon dioxide (CO2) gas at standard temperature and pressure (STP), we need to use the stoichiometry of the balanced equation for the reaction between sodium bicarbonate and acetic acid.
The balanced equation for the reaction between sodium bicarbonate and acetic acid is as follows:
NaHCO3 + CH3COOH → CO2 + H2O + NaCH3COO
From the balanced equation, we can see that one mole of sodium bicarbonate produces one mole of carbon dioxide. Therefore, we need to calculate the number of moles of CO2 to determine the amount of sodium bicarbonate required.
1 mole of gas at STP occupies 22.4 liters. Since we want to produce 0.25 L of CO2, the number of moles of CO2 can be calculated as follows:
0.25 L CO2 × (1 mole CO2 / 22.4 L CO2) = 0.01116 moles CO2
Since the stoichiometry of the equation indicates that one mole of NaHCO3 produces one mole of CO2, the number of moles of NaHCO3 needed is also 0.01116 moles.
To convert moles to grams, we need to use the molar mass of sodium bicarbonate, which is 84.01 g/mol.
0.01116 moles NaHCO3 × 84.01 g/mol = 0.938 g NaHCO3
Therefore, to produce 0.25 L of CO2 gas at STP with excess acetic acid, you would need to use approximately 0.938 grams of sodium bicarbonate (NaHCO3).
To determine the number of grams of sodium bicarbonate (NaHCO3) needed to produce 0.25 L of carbon dioxide (CO2) gas at standard temperature and pressure (STP), you would need to use the balanced chemical equation for the reaction between sodium bicarbonate and acetic acid.
The balanced chemical equation for the reaction is:
NaHCO3 + CH3COOH -> CO2 + H2O + NaCH3COO
In this equation, the ratio of NaHCO3 to CO2 is 1:1, meaning that for every 1 mole of sodium bicarbonate, 1 mole of carbon dioxide is produced.
To calculate the number of moles of carbon dioxide produced, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (STP = 1 atm)
V = volume (0.25 L)
n = number of moles of CO2
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature (STP = 273.15 K)
Rearranging the equation to solve for n, we have:
n = PV / RT
Substituting the given values, we have:
n = (1 atm) * (0.25 L) / (0.0821 L.atm/mol.K * 273.15 K)
n ≈ 0.011 mol
Since the ratio of NaHCO3 to CO2 is 1:1, the number of moles of sodium bicarbonate needed is also 0.011 mol.
To convert moles to grams, we need to use the molar mass of NaHCO3. The molar mass of NaHCO3 is:
Na = 23 g/mol
H = 1 g/mol
C = 12 g/mol
O = 16 g/mol (3 oxygen atoms in NaHCO3)
Molar mass of NaHCO3 = 23 + 1 + 12 + (16 * 3) = 84 g/mol
So, the number of grams of sodium bicarbonate needed is:
grams = moles * molar mass
grams = 0.011 mol * 84 g/mol
grams ≈ 0.924 g
Therefore, you would need to use approximately 0.924 grams of sodium bicarbonate to produce 0.25 L of CO2 gas at STP.