Molar Masses: Iron (II) ammonium sulfate hexahydrate:Fe(NH 4 ) 2 (SO 4 ) 2 x6H 2

Question

Molar Masses:
Iron (II) ammonium sulfate hexahydrate:Fe(NH4)2(SO4)2 x6H2O [limiting reagent]
Molar Mass: 392
Iron (II) oxalate dihydrate: FeC2O4 x2H2O
Molar Mass: 180
Oxalic acid: H2C2O4
Molar Mass: 90
Potassium trisoxalatoiron (III) trihydrate:K3Fe(C2O4)3 x3H2O
Molar Mass: 491
Potassium oxalate monohydrate:K2C2O4 x H2O
Molar Mass: 184

Question:
Suppose you start with 5.00 g ofFe(NH4)2(SO4)2 x6H2O and 30.0 mL of 1.0 MH2C2O4. What is the theoreticalyield of FeC2O4 x 2H2O (in grams)at the end of the precipitation reaction?

Explanation / Answer

Mass of Fe(NH4)2(SO4)2 .6H2O = 5.00 g Moles ofFe(NH4)2(SO4)2.6H2O = Mass / molar mass                                                     =5.00 g / 392 g/mol                                                    = 0.0127 moles Moles of H2C2O4 = Molarity *volume                            = 1.0 M * 0.03 L                            = 0.03 moles 1 mole ofFe(NH4)2(SO4)2.6H2O reacts to give 1 moleof  FeC2O4 x 2H2O So moles of  FeC2O4 x2H2O = 0.0127 moles So mass (theoretical yield ) = Moles * molar mass                                          =0.0127 * 184 g/mol                                          =2.33 g

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