What volume in mL of a 1.06 M NaOH solution is required to neutralize 12 g of HCl?
Here is a worked example of a stoichiometry problem. Just follow the steps. Remember M x L = moles.
http://www.jiskha.com/science/chemistry/stoichiometry.html
how much of 1n naoh are required to neutralize 0.73 gm of hcl
To determine the volume of a specific solution required for a chemical reaction, you can use the concept of stoichiometry and the equation:
M1V1 = M2V2
where M1 is the molarity of the first solution (NaOH), V1 is the volume of the first solution, M2 is the molarity of the second solution (HCl), and V2 is the volume of the second solution.
In this case, we are given the molarity of NaOH (1.06 M) and the mass of HCl (12 g). However, we need to convert the mass of HCl to its number of moles before calculating the volume.
1. Calculate the number of moles of HCl:
To do this, you need to know the molecular weight of HCl, which is approximately 36.5 g/mol.
Number of moles = mass / molecular weight
Number of moles = 12 g / 36.5 g/mol
2. Calculate the volume of NaOH solution:
Now we can use the equation mentioned earlier, M1V1 = M2V2, to find the volume of NaOH solution (V1).
Rearranging the equation, we have:
V1 = (M2V2) / M1
Given:
M1 = 1.06 M (molarity of NaOH)
M2 = 1 (since we are assuming the same volume of HCl as NaOH)
Substituting these values into the equation:
V1 = (1 x V2) / 1.06
Note: Since it is not specified what the volume of HCl (V2) is, we cannot determine the exact volume of NaOH solution required. However, we can still calculate the volume ratio between the two solutions.
Let's suppose that the volume of HCl (V2) is 100 mL.
V1 = (1 x 100 mL) / 1.06
V1 ≈ 94.34 mL
Therefore, approximately 94.34 mL of a 1.06 M NaOH solution is required to neutralize 12 g of HCl, assuming HCl has a volume of 100 mL.