Using the balanced equation of:
HCl+NaOH-->H20+NaCl, how many moles of HCl could be neutralized with 13.2 mL of 3.00 M NaOH?
mols NaOH = M x L = ?
mols HCl = mols NaOH
To find the number of moles of HCl that can be neutralized with 13.2 mL of 3.00 M NaOH, we need to use the stoichiometry of the balanced equation.
The balanced equation is:
HCl + NaOH -> H2O + NaCl
From the equation, we can see that the stoichiometric ratio of HCl to NaOH is 1:1. This means that for every 1 mole of HCl, 1 mole of NaOH is required for complete neutralization.
First, we need to find the number of moles of NaOH in 13.2 mL of 3.00 M NaOH. To do this, we use the formula:
moles = concentration (M) x volume (L)
Since the volume is given in mL, we need to convert it to liters:
Volume (L) = 13.2 mL / 1000 mL/L = 0.0132 L
Now we can calculate the moles of NaOH:
moles of NaOH = 3.00 M x 0.0132 L = 0.0396 moles
According to the stoichiometry of the balanced equation, this means that 0.0396 moles of HCl are required for complete neutralization.
Therefore, the number of moles of HCl that could be neutralized with 13.2 mL of 3.00 M NaOH is 0.0396 moles.