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. step 2: K3[Fe(c2O4)3].3H2O Q/ how do find the theoretical yield in mole and gram ? Molar mass.....491 g/mol.............. theoretical yield..................mol theoretical yield...........................g. mass of watch glass and K3[Fe(C2O4)3].3H2O........34.07 g..... mass of watch glass ...31.79.g........ mass of K3[Fe(C2O4)3].3H2O....2.28.g...... Q/ how do we find the yield of K3[Fe(C2O4)3].3H2O? % yield of K3[Fe(C2O4)3].3H2O..........................................? Q1/ using the moles of K3[Fe(C2O4)3].3H2O, calculate the theoretical yield, in grams, of K3[Fe(C2O4)3].3H2O. Q2/ show the calculation for the percent yield of K3[Fe(C2O4)3].3H2O using your actual yield and the theoretical yield.

Explanation / Answer

From your balanced equation: 1 mol FeSO4(NH4)2SO4.6H2O will produce 1mol FeC2O4 Molar mass FeSO4(NH4)2SO4.6H2O = 392.14 g/mol 4.0g = 4/392.14 = 0.0102 mol Theoretical production - 0.0102mol FeC2O4 Molar mass FeC2O4 = 143.86 g/mol 0.0102mol = 0.0102*143.86 = 1.46g Theoretical production of FeC2O4 = 1.46g

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