85.50 mL of 0.0815 M sodium hydroxide is reacted with 8.15 g oxalic acid as show

Question

85.50 mL of 0.0815 M sodium hydroxide is reacted with 8.15 g oxalic acid as shown in the reaction below. What mass of the excess reactant remains after the reaction is complete?
H2C2O4(s) + 2 NaOH(aq) ? Na2C2O4(aq) + 2 H2O(l)

Explanation / Answer

Grams of NaOH = 0.0815*85.5 = 6.96825 Grams (2+24+(4*16)) Grams of H2C2O4 reacts with 2*(23+16+1) Grams of NaOH =>90 Grams of H2C2O4 reacts with 80 Grams of NaOH =>8.15 Grams Reacts with (8.15/90)*80 = 7.244444444 < 6.96825 Grams So H2CO4 Remains. 80 Grams of NaOH reacts with 90 Grams of H2C2O4 =>6.96825 Grams of NaOH reacts with (6.96825/80)*90 = 7.83928125 Grams of Oxalic Acid mass of the excess reactant remains after the reaction is complete = 8.15-7.83928125 = 0.31071875 Grams of Oxalic Acid

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