How many milliliters of 0.05 N HC1 are required to neutralize exactly 8.0 g of NaOH???
how many moles of NaOH in 8g?
It will take that many moles of HCl
Now, how many ml of .05N HCl will that take?
how many moles of NaOH in 8g?
It will take that many moles of HCl
Now, how many ml of .05N HCl will that take?
To find out how many milliliters of 0.05 N HCl are required to neutralize 8.0 g of NaOH, we need to use the equation:
n(HCl) × V(HCl) = n(NaOH) × V(NaOH)
Where:
n(HCl) is the number of moles of HCl
V(HCl) is the volume of HCl in liters
n(NaOH) is the number of moles of NaOH
V(NaOH) is the volume of NaOH in liters
First, we need to calculate the number of moles of NaOH:
molar mass of NaOH = 22.99 g/mol (Na) + 16.00 g/mol (O) + 1.01 g/mol (H)
molar mass of NaOH = 39.99 g/mol + 16.00 g/mol + 1.01 g/mol = 56.00 g/mol
Therefore, the number of moles of NaOH is calculated as follows:
n(NaOH) = mass of NaOH / molar mass of NaOH
n(NaOH) = 8.0 g / 56.00 g/mol
n(NaOH) = 0.1429 mol
Now, we can substitute the values into the equation:
n(HCl) × V(HCl) = n(NaOH) × V(NaOH)
n(HCl) × 0.05 L = 0.1429 mol × V(NaOH)
Since we want to find the volume of HCl in milliliters, we need to convert liters to milliliters:
1 L = 1000 mL
Now we can calculate the volume of HCl:
n(HCl) × 50 mL = 0.1429 mol × V(NaOH)
n(HCl) × 50 mL = 0.1429 mol × 1000 mL
n(HCl) × 50 mL = 142.9 mol·mL
n(HCl) = 142.9 mol·mL / 50 mL
n(HCl) = 2.858 mol
Therefore, you would need approximately 2.858 milliliters of 0.05 N HCl to neutralize exactly 8.0 g of NaOH.
To determine the number of milliliters of 0.05 N HCl required to neutralize 8.0 g of NaOH, you need to follow a series of steps:
Step 1: Calculate the molar mass of NaOH.
The molar mass of NaOH can be obtained by using the atomic masses of each element: Na (Sodium) = 22.99 g/mol, O (Oxygen) = 16.00 g/mol, and H (Hydrogen) = 1.01 g/mol. Add these atomic masses together to find the molar mass of NaOH:
Na + O + H = 22.99 + 16.00 + 1.01 = 40.00 g/mol.
Step 2: Convert the mass of NaOH to moles.
Use the formula: moles = mass / molar mass
moles = 8.0 g / 40.00 g/mol = 0.20 mol
Step 3: Determine the stoichiometric ratio between NaOH and HCl.
From the balanced equation for the neutralization reaction between NaOH and HCl:
NaOH + HCl -> NaCl + H2O
We can see that the ratio is 1:1. This means that 1 mole of NaOH reacts with 1 mole of HCl.
Step 4: Calculate the volume of 0.05 N HCl required.
Since the volume is in milliliters, we need to convert the moles of NaOH to milliliters of HCl.
First, calculate the number of moles of HCl needed by multiplying the moles of NaOH by the stoichiometric ratio:
moles HCl = 0.20 mol NaOH
So, 0.20 moles of NaOH react with 0.20 moles of HCl.
Now, we need to convert the moles of HCl to milliliters using the concentration of HCl (0.05 N).
Remember that the term "N" stands for "normality," which can be converted to molarity (M) by dividing by the stoichiometric factor of the reaction, in this case, 1:
Molarity (M) = Normality (N) / 1
Therefore, the molarity (M) of 0.05 N HCl is 0.05 M.
Finally, use the formula: moles = volume (in liters) * molarity.
We need to convert milliliters to liters:
1 L = 1000 mL
So, 1 mL = 1/1000 L = 0.001 L
moles HCl = volume (in liters) * molarity
0.20 mol = volume (in liters) * 0.05 M
Solving for the volume:
volume (in liters) = 0.20 mol / 0.05 M = 4 L
But, we need the volume in milliliters:
volume (in milliliters) = volume (in liters) * 1000 mL/L
volume (in milliliters) = 4 L * 1000 mL/L = 4000 mL
Therefore, 4000 mL (or 4.0 L) of 0.05 N HCl is required to neutralize exactly 8.0 g of NaOH.