i PRODUCT K 3 CrC 2 O 4 3 3H 2 O Average mass of Na 2 C 2 O 4 02190 g 02029 g 2

I product k 3 crc 2 o 4 3 3h 2 o average mass of na 2

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i) PRODUCT = K 3 [Cr(C 2 O 4 ) 3 . 3H 2 O Average mass of Na 2 C 2 O 4 = 0.2190 g + 0.2029 g 2 = 0.2110g Number of moles of Na 2 C 2 O 4 = 0.2110g Na 2 C 2 O 4 x 1 molNa 2 C 2 O 4 134.0 g Na 2 C 2 O 4 = 1.5746 x10 -3 mol Na 2 C 2 O 4 Number of moles of KMnO 4 solution =1.567 x10 -3 mol Na 2 C 2 O 4 x 2 mol KMnO 4 5 mol Na 2 C 2 O 4 = 6.2985 x 10 -4 mol KMnO 4 Average volume of KMnO4 = 9.00 ml + 9.80 ml 2 = 9.40ml = 0.0094 L Concentration of KMnO 4 = 6.2985 x 10 4 mol KMnO 4 0.0094 L = 0.0670 M KMnO 4 Average volume of KMnO4 ¿ 4.30 ml + 4.30 ml 2 = 4.30ml = 0.0043 L Number of moles of KMnO4
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=0.0670 M x 0.0043 L = 2.8810 x10-4 mol KMnO4 Number of moles of MnO4- 2.8810 x10-4 mol KMnO4 x 1 mol MnO 4 ¿ 1 mol KMnO 4 ¿ = 2.8810 x10-4 mol MnO4- Number of moles of C2O4 2- 2.8810 x10-4 mol MnO4- x 2 mol MnO 4 ¿ 5 molC 2 O 42 ¿ ¿ ¿ = 7.2025 x 10-4 mol C2O4 2- Mass of oxalate = 7.2025 x 10-4 mol C2O4 2- x 88.0 g mol-1 = 0.0634 g C2O4 2- Percent of oxalate = 0.0634 gC 2 O 42 ¿ 0.2024 g ¿ X 100% = 31.32% Theoretical percentage oxalate in complex = 3 x 88.0 gmol 1 487.36 g mol 1 x 100% = 54.17% Percentage purity of complex = 31.32% 54.17 % x 100% =57.82% ii) PRODUCT = K 2 [Cu(C 2 O 4 ) 2 (H 2 O) 2 ]
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Concentration of KMnO 4 = 6.2985 x 10 4 mol KMnO 4 0.0094 L = 0.0670 M KMnO 4 Average volume of KMnO4 ¿ 6.90 ml + 7.20 ml 2 = 7.05ml = 0.00705 L Number of moles of KMnO4 =0.0670 M x 0.00705 L = 4.7235 x10-4 mol KMnO4 Number of moles of MnO4- = 4.7235 x10-4 mol KMnO4 x 1 mol MnO 4 ¿ 1 mol KMnO 4 ¿ = 4.7235 x10-4 mol MnO4- Number of moles of C2O4 2- = 4.7235 x10-4 mol MnO4- x 2 mol MnO 4 ¿ 5 molC 2 O 42 ¿ ¿ ¿ = 1.1809x10-3 mol C2O4 2- Mass of oxalate = 1.1809x10-3 mol C2O4 2- x 88.0 g mol-1 = 0.1039 g C2O4 2- Percent of oxalate = 0.1039 gC 2 O 42 ¿ 0.2089 g ¿ X 100% = 49.74% Theoretical percentage oxalate in complex = 3 x 88.0 gmol 1 487.36 g mol 1 x 100% = 54.17%
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Percentage purity of complex = 49.74% 54.17% x 100% =91.82% iii) PRODUCT = K 3 [Fe(C 2 O 4 ) 3 ].3H 2 O Concentration of KMnO 4 = 6.2985 x 10 4 mol KMnO 4 0.0094 L = 0.0670 M KMnO 4 Average volume of KMnO4 ¿ 7.00 ml + 7.40 ml 2 = 7.20ml = 0.0072 L Number of moles of KMnO4 =0.0670 M x 0.0072 L = 4.8240 x10-4 mol KMnO4 Number of moles of MnO4- = 4.8240 x10-4 mol KMnO4 x 1 mol MnO 4 ¿ 1 mol KMnO 4 ¿ = 4.8420 x10-4 mol MnO4- Number of moles of C2O4 2- = 4.8420 x10-4 mol MnO4- x 2 mol MnO 4 ¿ 5 molC 2 O 42 ¿ ¿ ¿ = 1.2060 x10-3 mol C2O4 2- Mass of oxalate = 1.2060 x10-3 mol C2O4 2- x 88.0 g mol-1 = 0.1061 g C2O4 2-
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Percent of oxalate = 0.1061 gC 2 O 42 ¿ 0.2074 g ¿ X 100% = 51.16% Theoretical percentage oxalate in complex = 3 x 88.0 gmol 1 487.36 g mol 1 x 100% = 54.17% Percentage purity of complex = 51.16% 54.17 % x 100% =94.44% 3) Predict and draw the structure of the synthesize compound For K 3 [Cr(C 2 O 4 ) 3 . 3H 2 O,
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For K 2 [Cu(C 2 O 4 ) 2 (H 2 O) 2 ]
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For K 3 [Fe(C 2 O 4 ) 3 ].3H 2 O DISCUSSION:
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In this experiment, we have to do titration method. We have to use KMnO4 solution for the titration. The purpose of this experiment is to determine the oxalate content in coordination compound. The experiment is divided into parts, part A and B; where each part discussed each principle. Standardization process is defined as “the process of determining the exact concentration (molarity) of a solution. Titration is one type of analytical procedure often used in standardization. In a titration, an exact volume of one substance is reacted with a known amount of another substance. For part A, we have to use dried Na2C2O4 and
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