I cant figure out what Im doing wrong **Text Transcription for Educational Website:**---A solution of oxalic acid dihydrate (H₂C₂O₄·2H₂O) with a known concentration of **0.400 M H₂C₂O₄·2H₂O** is titrated with a **0.333 M NaOH** solution. How many **L NaOH** are required to reach the second equivalence point with a starting volume of **65.0 mL H₂C₂O₄·2H₂O**, according to the following balanced chemical equation:\[ \text{H₂C₂O₄·2H₂O} + 2 \text{NaOH} \rightarrow \text{Na₂C₂O₄} + 4\text{H₂O} \]**Calculation Steps:**1. **Convert Volume to Moles:** \[ \frac{\text{mol H₂C₂O₄·2H₂O}}{\text{L H₂C₂O₄·2H₂O}} \times \frac{\text{mL H₂C₂O₄·2H₂O}}{1} = \frac{65.0 \text{ mL H₂C₂O₄·2H₂O}}{1 \text{ mol H₂C₂O₄·2H₂O}} \]2. **Use Stoichiometry to Convert to NaOH:** \[ \times \frac{2 \text{ mol H₂C₂O₄·2H₂O}}{0.333 \text{ mol NaOH}} \]3. **Calculate Final Volume:** \[ = 0.156 \text{ L NaOH} \]**Diagram Explanation:**- The top section describes the titration process and provides the balanced chemical equation.- Below this, a step-by-step calculation is shown, converting volume to moles using molarity, then applying stoichiometry to find the volume of NaOH.- At the bottom, there is a toolbox with various numerical and unit conversion factors, including buttons for molarity, volume, and mole conversion related to both H₂C₂O₄·2H₂O and NaOH.This setup demonstrates the systematic approach to solving a titration problem using molarity and sto

Given that : Concentration of H2C2O4.2H2O=0.400 MConcnetration of NaOH=0.333 MThe volume of H2C2O4.2H2O used=65.0 mL The balanced equation for the reaction is : H2C2O4·2H2O + 2NaOH → Na2C2O4 + 4H2ONow, the number of moles of H2C2O4.2H2O can be calculated as : Number of moles of H2C2O4·2H2O=Volume (in L)×Concentration65mL H2C2O4·2H2O ×1L H2C2O4·2H2O1000mL H2C2O4·2H2O×0.400mol H2C2O4·2H2O1L H2C2O4·2H2O ⇒0.026 mol H2C2O4·2H2O Now, according to the balanced equation, 1 mole of H2C2O4.2H2O requires 2 moles of NaOH. So, 0.026 moles of H2C2O4.2H2O will require : =0.026mol H2C2O4.2H2O×2mol NaOH1mol H2C2O4.2H2O⇒0.052 mol NaOHNow, the volume of NaOH required can be calculated as : Volume=Number of moles of NaOHConcentration of NaOH=0.052 mol NaOH0.333 mol NaOH×1 L NaOH⇒0.156L NaOH Thus, the overall calculation can be given as : 65 mL H2C2O4·2H2O ×1L H2C2O4·2H2O1000 mL H2C2O4·2H2O×0.400 mol H2C2O4·2H2O1L H2C2O4·2H2O ×2 mol NaOH1 mol H2C2O4.2H2O×1L NaOH0.333 mol NaOH⇒0.156 L NaOH

Science

Chemistry

A solution of oxalic acid dihydrate (H:C2O4-2H:O) with a known concentration of 0.400 M H:C2O4-2H20 is titrated with a 0.333 M NaOH solution. How many L NaOH are required to reach the second equivalence point with a starting volume of 65.0 mL H:C2O4-2H2O , according to the following balanced chemical equation: H:C:O4 2H:0 + 2 NaOH – Na:C:O4 + 4 H:O mol H.GO. ZH:O 65.0 mL H:C2O4-2H2O 2 mol H:C2O4 2H:0 0.156 L NAOH LH:C:04-2H:0 1 molH.GO. ZH:O 0.333 mol NaOH -O.-2H:0 65.0 mL H:C:0.-2H:0 2 mol H:C:O.-2H:0 = 0.156 L NaOH 2H:0 1 mol H:C:O.-2H:0 0.333 mol NaOH ADD FACTOR DELETE ANSWER RESET *( ) 0.400 0.156 2 0.001 156 7.81 x 10 0.156 0.0781 0.333 4 39.0 78.1 1.56 x 10 1000 0.0520 5.20 x 10° 65.0 mol H:C204-2H:O g NaOH mL NaOH MH:C2O+-2H2O g H:C2O+-2H2O mol NaOH L NaOH M NaOH L H:C:O.-2H:O mL H:C2O4-2H:0