98. You have 0.554g sample that is a mixture of oxalic acid, H2C2O4 and another

Question

98. You have 0.554g sample that is a mixture of oxalic acid, H2C2O4 and another solid that does not react with sodium hydroxide. If 29.58 mL of 0.550 M NaOH is required to titrate the oxalic acid in the 0.554g sample to the equivalence point, what is the weight percent of oxalic acid in the mixture? Oxalic acid and NaOH react according to the equation H2C2O4 (aq) + 2 NaOH(aq)--->NaC2O4(aq) + 2 H2O (l)

Explanation / Answer

H2C2O4 (aq) + 2 NaOH(aq)--->NaC2O4(aq) + 2 H2O (l),,,,,,,,,, moles of NaOH =29.58 mL *0.550 /1000 = 0.0162 mol of NaOH,,,,, moles of Oxalic acid = moles of NaOH/2 [from equation] =0.00813 mol of H2C2O4,,,,,,,, grams of H2C2O4 = 0.0183*90.035 = 0.7324 gm,,,,,,,,,, percent oxalic acid =[0.7324 g/4.554g]*100 % = 16.1% H2C2O4

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