Question

Consider the following two reactions involving oxalic acid. 2Mn04- + 5H2C204 + 6H+ + 2Mn2+ + 10002 + 8H20 H2C2O4 + 204- → C2042- + 2H20 What volume of 0.0100 M KMnO4 solution will titrate a solution prepared from 37.0 mg of H2C204? 2.67 mg Submit Answer Incompatible units. No conversion found between "mg" and the required units. Tries 0/5 Previous Tries What volume of 0.0100 M NaOH solution will titrate a solution prepared from 37.0 mg of H2C204? Submit Answer Tries 0/5
please can you help me

Answer 1

Comment in case of any doubt.

→ 37.0 my Holz Oy = 37.0X1669 HQ Oy x Imel Hab2O4 قامی مہور مه جس 90 . 03 = 4.1097 xio' mol H2O4 KMnOy moles needed = amo kinay O x 4.1097x10molt goy 5 mol HG OY 1 = 1.644x10 mol Now, Mamnou x VKM nou = moles of KMn Oyneeded =) 0.0100 MXV Anna la 1.640 x10 mah izred on =) Vamnou= 0.016442 = (16.4mL NaOH males needed = and No OH x 1.64 UXTO molt ca. 1 Imal gott sme 76204 = 3.288xjömed Mwadh* Von = males of NaOH cut browser yma I needed 20-0100Mxv 13-2888100 ml => Naoke= 3.288x16?L= 32.9 mb] → The first end point cresol purple occurs around 1. pis, at which the would have been neutralized along with first and point of Hapoy. If ni be moes He and he be moles of H₂ Poy initially. Then ni tha= moles of NaOH used at 36.94mL = 0.03694 LX01M 1. =) nt m = 0.003694 mal At and end point (thinal blue both end points would have reached of H₂ Poy. That is and and point is reached when exetra Nool useal = 56.28 - 36.94 m2 = 19.34 ML = 0.01934L So, n₂ = 0.1MX0.01934 L = 0.001934 mol = 1.93 mmol Then, ni=0.003694 -0.001934= 0.00176 mol = 1:76mmal Thus He present initially = 11:76mmal) H₃ Pa, present = 11.93 mmol "Naom