Calculate the mass of baking soda, NaHCO3, needed to neutralize 3.0 mL of 3.0 M hydrochloric acid, HCl(aq).
Ok, so I'm super confused. I have the equation, but I'm stuck with what to do with it.
Equation:
NaHCO3 + HCL --> NaCl + H2O + CO2
I don't think that you need the equation- just calculate the moles in 3.0ml of a 3.0M solution of HCl and then calculate the mass needed from the number of moles for molecular weight of NaHCO3.
That makes so much more sense!! Thanks :)
To calculate the mass of baking soda (NaHCO3) needed to neutralize the hydrochloric acid (HCl), we can use stoichiometry to relate the moles of HCl to the moles of NaHCO3.
Here's how you can approach the problem step-by-step:
Step 1: Determine the balanced chemical equation.
The balanced equation is:
NaHCO3 + HCl → NaCl + H2O + CO2
Step 2: Convert the volume of the hydrochloric acid to moles.
To do this, we need to use the concentration of the hydrochloric acid.
Molarity (M) = moles of solute / liters of solution
Given:
Volume of HCl = 3.0 mL = 3.0 * 10^(-3) L (converted to liters)
Molarity of HCl = 3.0 M
Using the equation M = moles of solute / liters of solution, we can rearrange it to solve for moles of solute:
moles of solute = Molarity * liters of solution
moles of HCl = 3.0 M * 3.0 * 10^(-3) L = 9.0 * 10^(-3) mol
Step 3: Use stoichiometry to determine the moles of NaHCO3 needed.
From the balanced chemical equation, we can see that the stoichiometric ratio of HCl to NaHCO3 is 1:1. Therefore, the moles of NaHCO3 needed is also 9.0 * 10^(-3) mol.
Step 4: Convert moles of NaHCO3 to mass.
To convert moles to mass, we need to use the molar mass of NaHCO3.
The molar mass of NaHCO3 = (22.99 g/mol + 1.01 g/mol + 12.01 g/mol + 3 * 16.00 g/mol) = 84.01 g/mol
mass of NaHCO3 = moles of NaHCO3 * molar mass of NaHCO3
mass of NaHCO3 = 9.0 * 10^(-3) mol * 84.01 g/mol
mass of NaHCO3 = 0.756 g
Therefore, approximately 0.756 grams of baking soda (NaHCO3) is needed to neutralize 3.0 mL of 3.0 M hydrochloric acid (HCl).
To solve this problem, you need to use stoichiometry and the concept of molarity.
First, write down the balanced chemical equation:
NaHCO3 + HCl -> NaCl + H2O + CO2
Next, determine the stoichiometry of the reaction. From the balanced chemical equation, you can see that 1 mol of NaHCO3 reacts with 1 mol of HCl.
Now, use the given information to calculate the number of moles of HCl in 3.0 mL of 3.0 M HCl solution.
To do this, you need to convert the volume of the solution to moles using the molarity (M) of the solution. The molarity is given as 3.0 M, which means there are 3.0 moles of HCl in 1 liter (1000 mL) of the solution.
Using the formula:
moles = volume (in liters) x molarity
Convert the volume to liters:
Volume = 3.0 mL = 3.0 x 10^-3 L
Now, calculate the moles of HCl:
moles = 3.0 x 10^-3 L x 3.0 M = 9.0 x 10^-3 moles
Since the stoichiometric ratio is 1:1, the number of moles of NaHCO3 needed is also 9.0 x 10^-3 moles.
Finally, calculate the mass of NaHCO3:
To determine the mass of NaHCO3, you need to use its molar mass, which can be found on the periodic table.
NaHCO3:
1 Na = 22.99 g/mol
1 H = 1.01 g/mol
1 C = 12.01 g/mol
3 O = 16.00 g/mol (each)
Adding up the molar masses:
22.99 + 1.01 + 12.01 + (3 x 16.00) = 84.01 g/mol
Now, use the equation:
mass = moles x molar mass
Substitute the values:
mass = 9.0 x 10^-3 moles x 84.01 g/mol
Calculating the mass:
mass = 0.756 g
Therefore, you would need approximately 0.756 grams of baking soda (NaHCO3) to neutralize 3.0 mL of 3.0 M hydrochloric acid (HCl(aq)).