how many grams of oxalic acid dihydrate to completely neutralize 25ml of 0.12 M of sodium hydroxide
H2C2O4 + 2NaOH ==> Na2C2O4 + 2H2O
moles NaOH = M x L = ??
moles H2C2O4 = 1/2 moles NaOH.
grams H2C2O4.H2O = moles x molar mass.
To calculate the grams of oxalic acid dihydrate required to completely neutralize 25 ml of 0.12 M sodium hydroxide, we can use the concept of stoichiometry. The balanced chemical equation for the reaction between oxalic acid dihydrate (H2C2O4·2H2O) and sodium hydroxide (NaOH) is:
H2C2O4·2H2O + 2NaOH -> Na2C2O4 + 4H2O
From the balanced equation, we can see that 1 mole of oxalic acid dihydrate reacts with 2 moles of sodium hydroxide. Therefore, we need to determine the number of moles of sodium hydroxide in 25 ml of 0.12 M solution.
Step 1: Calculate the number of moles of sodium hydroxide:
Moles of NaOH = Volume (in liters) x Concentration (in M)
Moles of NaOH = (25 ml / 1000 ml) x 0.12 M
Moles of NaOH = 0.003 moles
Step 2: Use the stoichiometry to find the moles of oxalic acid dihydrate:
From the balanced equation, we can see that 1 mole of oxalic acid dihydrate reacts with 2 moles of sodium hydroxide.
Therefore, the moles of oxalic acid dihydrate needed = 0.003 moles x (1 mole / 2 moles)
Moles of oxalic acid dihydrate needed = 0.0015 moles
Step 3: Calculate the grams of oxalic acid dihydrate required:
To find the grams of oxalic acid dihydrate, we need to multiply the moles by its molar mass. The molar mass of oxalic acid dihydrate is 126.07 g/mol.
Grams of oxalic acid dihydrate = Moles of oxalic acid dihydrate x Molar mass of oxalic acid dihydrate
Grams of oxalic acid dihydrate = 0.0015 moles x 126.07 g/mol
Grams of oxalic acid dihydrate = 0.1891 grams
Therefore, approximately 0.1891 grams of oxalic acid dihydrate is required to completely neutralize 25 ml of 0.12 M sodium hydroxide.
To determine the number of grams of oxalic acid dihydrate needed to completely neutralize 25 ml of 0.12 M sodium hydroxide, we can use the concept of stoichiometry. The balanced chemical equation for the reaction between oxalic acid dihydrate (H2C2O4·2H2O) and sodium hydroxide (NaOH) is:
H2C2O4·2H2O + 2NaOH -> Na2C2O4 + 4H2O
From the balanced equation, we can see that the mole ratio between oxalic acid dihydrate and sodium hydroxide is 1:2. This means that for every 1 mole of oxalic acid dihydrate (H2C2O4·2H2O), we need 2 moles of sodium hydroxide (NaOH) to completely neutralize the solution.
First, we need to determine the number of moles of sodium hydroxide present in the solution. We know that the concentration (Molarity) of the sodium hydroxide solution is 0.12 M (mol/L), and the volume is 25 ml (0.025 L). We can calculate this as:
moles of NaOH = concentration x volume
moles of NaOH = 0.12 M x 0.025 L
moles of NaOH = 0.003 moles
Now that we know the number of moles of sodium hydroxide, we can determine the number of moles of oxalic acid dihydrate required. Since the mole ratio between the two is 1:2, we can calculate this as:
moles of H2C2O4·2H2O = 0.003 moles x 1/2
moles of H2C2O4·2H2O = 0.0015 moles
Finally, we need to convert the moles of oxalic acid dihydrate into grams. The molar mass of oxalic acid dihydrate (H2C2O4·2H2O) is calculated as:
molar mass of H2C2O4·2H2O = (2 x molar mass of H) + (2 x molar mass of C) + (4 x molar mass of O) + (2 x molar mass of H2O)
molar mass of H2C2O4·2H2O = (2 x 1.008 g/mol) + (2 x 12.01 g/mol) + (4 x 16.00 g/mol) + (2 x 18.02 g/mol)
molar mass of H2C2O4·2H2O = 126.07 g/mol
Then, we can calculate the grams of oxalic acid dihydrate as:
grams = moles x molar mass
grams of H2C2O4·2H2O = 0.0015 moles x 126.07 g/mol
grams of H2C2O4·2H2O = 0.1891 g
Therefore, approximately 0.1891 grams of oxalic acid dihydrate is needed to completely neutralize 25 ml of 0.12 M sodium hydroxide.