Sodium hydroxide reacts with oxalic acid as shown in the reaction below: H2C2O4(s) + 2 NaOH(aq) → Na2C2O4(aq) + 2 H2O(l)

Sodium hydroxide reacts with oxalic acid as shown in the reaction below: H2C2O4(s) + 2 NaOH(aq) → Na2C2O4(aq) + 2 H2O(l) a) If 0.312 moles of sodium hydroxide are reacted with 4.59 g of oxalic acid, what mass of the excess reactant remains after the reaction is complete? Be clear about which species you are calculating. b) The percent yield of this reaction is found to be 88.6%. If you require 4.65 moles of sodium oxalate for a subsequent reaction, how many kilograms of oxalic acid must be added to excess sodium hydroxide?

The Correct Answer and Explanation is:

a) Limiting Reactant and Excess Mass Remaining

Balanced reaction:H2C2O4+2NaOH→Na2C2O4+2H2O\text{H}_2\text{C}_2\text{O}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{C}_2\text{O}_4 + 2\text{H}_2\text{O}H2​C2​O4​+2NaOH→Na2​C2​O4​+2H2​O

Given:

• 0.312 mol NaOH
• 4.59 g oxalic acid

Molar mass of oxalic acid (H₂C₂O₄):
C: 12.01 × 2 = 24.02
H: 1.008 × 2 = 2.016
O: 16.00 × 4 = 64.00
Total = 90.03 g/mol

Moles of oxalic acid:4.59 g90.03 g/mol=0.0510 mol\frac{4.59\ \text{g}}{90.03\ \text{g/mol}} = 0.0510\ \text{mol}90.03 g/mol4.59 g​=0.0510 mol

From the stoichiometry:
1 mol oxalic acid reacts with 2 mol NaOH
So, 0.0510 mol oxalic acid requires:0.0510×2=0.102 mol NaOH0.0510 \times 2 = 0.102\ \text{mol NaOH}0.0510×2=0.102 mol NaOH

We have 0.312 mol NaOH, which is more than required.
Thus, oxalic acid is the limiting reactant, and NaOH is in excess.

NaOH used: 0.102 mol
NaOH remaining:0.312−0.102=0.210 mol0.312 – 0.102 = 0.210\ \text{mol}0.312−0.102=0.210 mol

Mass of excess NaOH:
Molar mass of NaOH = 22.99 + 15.999 + 1.008 = 39.997 g/mol0.210×39.997=8.40 g0.210 \times 39.997 = 8.40\ \text{g}0.210×39.997=8.40 g

Answer (a):
Excess species: Sodium hydroxide
Mass remaining after reaction: 8.40 g

b) Required Oxalic Acid for 4.65 mol Na₂C₂O₄ with 88.6% yield

From the balanced equation:
1 mol Na₂C₂O₄ is produced from 1 mol oxalic acid

So ideally, to get 4.65 mol of Na₂C₂O₄:4.65 mol H2C2O4required(theoretical)4.65\ \text{mol H}_2\text{C}_2\text{O}_4 required (theoretical)4.65 mol H2​C2​O4​required(theoretical)

Since the reaction is only 88.6% efficient, use:Actual mol needed=4.650.886≈5.25 mol\text{Actual mol needed} = \frac{4.65}{0.886} \approx 5.25\ \text{mol}Actual mol needed=0.8864.65​≈5.25 mol

Mass of oxalic acid needed:5.25 mol×90.03 g/mol=472.66 g5.25\ \text{mol} \times 90.03\ \text{g/mol} = 472.66\ \text{g}5.25 mol×90.03 g/mol=472.66 g

Convert to kilograms:472.66 g=0.4727 kg472.66\ \text{g} = 0.4727\ \text{kg}472.66 g=0.4727 kg

Answer (b):
You must add 0.4727 kilograms of oxalic acid to excess NaOH to obtain 4.65 mol of sodium oxalate at 88.6% yield.

This problem demonstrates how stoichiometric coefficients and percent yield guide reagent requirements and identify leftover materials. The limiting reactant determines the reaction extent, while excess reactants help ensure completion. Percent yield corrects for real-life inefficiencies.