How many grams of oxalic acid dihydrate (H2C2O4) are needed to react with 38.56 ml of 0.2500 sodium hydorxide

To determine the number of grams of oxalic acid dihydrate (H2C2O4) needed to react with 38.56 ml of 0.2500M sodium hydroxide (NaOH), we can follow these steps:

Step 1: Write the balanced chemical equation for the reaction between oxalic acid dihydrate (H2C2O4) and sodium hydroxide (NaOH).
H2C2O4 + 2 NaOH -> Na2C2O4 + 2 H2O

Step 2: Determine the mole-to-mole ratio between H2C2O4 and NaOH from the balanced equation.
From the balanced equation, we can see that the mole ratio between H2C2O4 and NaOH is 1:2. This means that for every 1 mole of H2C2O4, we need 2 moles of NaOH.

Step 3: Convert the volume of NaOH to moles.
Using the given concentration of NaOH (0.2500M) and the volume (38.56 ml), we can calculate the number of moles.
moles of NaOH = concentration * volume
moles of NaOH = 0.2500 mol/L * 0.03856 L
moles of NaOH = 0.00964 mol

Step 4: Convert moles of NaOH to moles of H2C2O4 using the mole-to-mole ratio.
Since the ratio of H2C2O4 to NaOH is 1:2, we can calculate the moles of H2C2O4 as follows:
moles of H2C2O4 = 0.00964 mol * (1 mol H2C2O4 / 2 mol NaOH)
moles of H2C2O4 = 0.00482 mol

Step 5: Convert moles of H2C2O4 to grams.
The molar mass of H2C2O4 is 90.03 g/mol (2*12.01 + 2*1.01 + 4*16.00).
grams of H2C2O4 = moles of H2C2O4 * molar mass of H2C2O4
grams of H2C2O4 = 0.00482 mol * 90.03 g/mol
grams of H2C2O4 = 0.434 g

Therefore, approximately 0.434 grams of oxalic acid dihydrate (H2C2O4) are needed to react with 38.56 ml of 0.2500M sodium hydroxide (NaOH).

To determine the number of grams of oxalic acid dihydrate (H2C2O4) needed to react with 38.56 mL of 0.2500 M sodium hydroxide (NaOH), we need to use stoichiometry, which involves considering the molar ratios between the reactants.

Here's the step-by-step process to find the answer:

Step 1: Determine the balanced chemical equation for the reaction between oxalic acid dihydrate (H2C2O4) and sodium hydroxide (NaOH). The balanced equation is:

H2C2O4 + 2 NaOH → Na2C2O4 + 2 H2O

This equation shows that for every 1 mole of oxalic acid dihydrate (H2C2O4), 2 moles of sodium hydroxide (NaOH) react.

Step 2: Calculate the number of moles of sodium hydroxide (NaOH) using its volume and molarity.
- Convert the given volume from milliliters (mL) to liters (L): 38.56 mL ÷ 1000 = 0.03856 L
- Use the formula: moles = volume (L) × molarity (M). Thus, the moles of NaOH = 0.03856 L × 0.2500 M = 0.00964 moles

Step 3: Apply the stoichiometry of the balanced equation to determine the moles of oxalic acid dihydrate (H2C2O4) needed.
- From the balanced equation, we see that 1 mole of H2C2O4 reacts with 2 moles of NaOH.
- So, the moles of H2C2O4 = (moles of NaOH) / 2 = 0.00964 moles / 2 = 0.00482 moles

Step 4: Calculate the molar mass of oxalic acid dihydrate (H2C2O4).
- H2C2O4: 2(1.00784 g/mol) + 2(12.01074 g/mol) + 4(15.999 g/mol) = 90.0349 g/mol
- The molar mass of H2C2O4 is approximately 90.0349 g/mol.

Step 5: Calculate the mass of oxalic acid dihydrate (H2C2O4) needed in grams.
- The mass of H2C2O4 = moles of H2C2O4 × molar mass of H2C2O4
- Mass of H2C2O4 = 0.00482 moles × 90.0349 g/mol = 0.4359 grams

Therefore, approximately 0.4359 grams of oxalic acid dihydrate (H2C2O4) are needed to react with 38.56 mL of 0.2500 M sodium hydroxide (NaOH).