Calculate the mass of sodium bicarbonate needed to neutralize 3.0 mL of 3.0 M HCl. Show your work.
NaHCO3 + HCl ==> NaCl + H2O + CO2
moles HCl = M x L = ??
moles NaHCO3 (use the coefficients in the balanced equation) = moles HCl
moles NaHCO3 = grams/molar mass. Solve for grams.
To calculate the mass of sodium bicarbonate needed to neutralize the given amount of HCl, we need to follow these steps:
1. Determine the moles of HCl: Start by calculating the moles of HCl using the given volume and molarity. We can use the formula:
Moles of solute = Volume of solution (in liters) × Molarity
The given volume is 3.0 mL, which needs to be converted to liters (1 mL = 0.001 L). The molarity is 3.0 M. Substituting these values into the formula:
Moles of HCl = 3.0 mL × 0.001 L/mL × 3.0 mol/L = 0.009 mol
Therefore, we have 0.009 moles of HCl.
2. Determine the stoichiometry: The balanced chemical equation for the reaction between HCl and sodium bicarbonate (NaHCO3) is:
HCl + NaHCO3 → NaCl + CO2 + H2O
From this equation, we can see that one mole of HCl reacts with one mole of NaHCO3.
3. Calculate the moles of sodium bicarbonate: Since the stoichiometry between HCl and NaHCO3 is one-to-one, the moles of NaHCO3 required will be the same as the moles of HCl. Therefore, we need 0.009 moles of NaHCO3.
4. Determine the molar mass of sodium bicarbonate: The molar mass of NaHCO3 can be calculated by adding up the atomic masses of each element in the compound. The atomic masses are as follows:
Na: 22.99 g/mol
H: 1.01 g/mol
C: 12.01 g/mol
O: 16.00 g/mol (three oxygen atoms in NaHCO3)
Molar mass of NaHCO3 = (22.99 g/mol) + (1.01 g/mol) + (12.01 g/mol) + (16.00 g/mol × 3) = 84.01 g/mol
5. Calculate the mass of sodium bicarbonate: Multiply the moles of NaHCO3 by its molar mass to get the mass.
Mass = Moles × Molar mass
Mass = 0.009 mol × 84.01 g/mol ≈ 0.76 g
Therefore, approximately 0.76 grams of sodium bicarbonate is needed to neutralize 3.0 mL of 3.0 M HCl.