FeSo4(nh4)2So.6h2o H2C2O4.2H2O Molar mass......392.14.......g/mol...... .. Molar

Question

FeSo4(nh4)2So.6h2o H2C2O4.2H2O Molar mass......392.14.......g/mol...... .. Molar mass...126.06 g/mol........ .... Mass.........2.57g.......... .... Mass..............1.31g.......... ..... Moles.......0.00655 mol....... .... Moles.............1.04*10^-2 mol. Q/ how do we find the limiting reagent, and the theoretical yield of FeC2O4.2H2O ? what is the limiting reagent?............................... theoretical yield of FeC2O4.2H2O ....................... mol.

Explanation / Answer

FeSO4·(NH4)2SO4· 6H2O + H2C2O4· 2H2O ---> FeC2O4.2H2O + (NH4)2SO4 + H2SO4 + 6H2O moles of FeSO4·(NH4)2SO4· 6H2O = 2.57/392.14 = 0.006554 moles of H2C2O4· 2H2O = 1.31/126.06 = 0.01039 1 mole FeSO4·(NH4)2SO4· 6H2O reacts with 1 mole of other reagent ; and FeSO4·(NH4)2SO4· 6H2O is less in quantity ... so it is limiting reagent Limiting Reagent : FeSO4·(NH4)2SO4· 6H2O because it will be used up completely ; 1 mole FeSO4·(NH4)2SO4· 6H2O gives 1 mole FeC2O4 .2H2O so 0.00655 moles will give 0.00655 moles FeC2O4.2H2O so mass = 0.00655 x 179.8951 = 1.179 gms -- theorotical yield

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