How many milliliters of 1.5 M NaHCO3 would it take to neutralize 300 mg of Benzoic Acid (C7H6O2)
mol benzoic acid = grams/molar mass
mols NaHCO3 = same
M NaHCO3 = mols NaHCO3/L NaHCO3. Solve for L and convert to mL.
Thanks for the help!
To determine the number of milliliters of 1.5 M NaHCO3 required to neutralize 300 mg of Benzoic Acid (C7H6O2), we first need to calculate the number of moles of Benzoic Acid and then use stoichiometry to find the corresponding moles of NaHCO3. Finally, we can convert the moles of NaHCO3 to volume in milliliters.
1. Calculate the number of moles of Benzoic Acid (C7H6O2):
- The molar mass of Benzoic Acid (C7H6O2) is calculated by adding up the atomic masses of each element: C(12.01 g/mol) + H(1.01 g/mol) + O(16.00 g/mol) + O(16.00 g/mol) = 122.12 g/mol
- Convert the given mass of Benzoic Acid to moles using the molar mass:
300 mg * (1 g / 1000 mg) * (1 mol / 122.12 g) = 0.002455 mol
2. Use the balanced chemical equation between Benzoic Acid (C7H6O2) and NaHCO3 to determine the stoichiometric ratio between the two compounds. The balanced equation is:
C7H6O2 + NaHCO3 -> C6H5COONa + H2O + CO2
Looking at the equation, we see that the stoichiometric ratio between Benzoic Acid and NaHCO3 is 1:1.
3. Now that we know the number of moles of Benzoic Acid, we can calculate the number of moles of NaHCO3 required:
- Since the stoichiometric ratio is 1:1, the moles of NaHCO3 required are also 0.002455 mol.
4. Finally, we can convert the moles of NaHCO3 to milliliters using the molarity (concentration) of NaHCO3:
- The concentration of the NaHCO3 solution is given as 1.5 M, which means it contains 1.5 moles of NaHCO3 per liter of solution.
- Convert moles to liters:
0.002455 mol * (1 L / 1.5 mol) = 0.001637 L
- Convert liters to milliliters:
0.001637 L * (1000 mL / 1 L) = 1.637 mL
Therefore, it would take approximately 1.637 milliliters of 1.5 M NaHCO3 to neutralize 300 mg of Benzoic Acid (C7H6O2).