Manganese(III) fluoride, MnF3, can be prepared by the following reaction: 2MnI2(

Question

Manganese(III) fluoride, MnF3, can be prepared by the following reaction: 2MnI2(s) + 13F2(g) ? 2MnF3(s) + 4IF5(l) If the percentage yield of MnF3 is always approximately 51%, how many grams of MnF3 can be expected if 25.0 grams of each reactant is used in an experiment?

Explanation / Answer

follow this 2 MnI2 + 13F2 -----> 2 MnF3 + 4 IF5 determine limiting reactant- MnI2: 45 g/ 308.75 g/mole MnI2= 0.146 moles MnI2 F2: 45 g/ 38 g/mole F2= 1.184 moles F2 from the bal. rxn., 6.5 moles of F2 per mole MnI2 is required, therfore MnI2 is limiting reactant from bal. rxn., 2 moles MnI2 reacts to form 1 mole MnF3 (1:1) 0.146 moles MnI2 x 111.93 g/mole MnF3 x 0.53(%yield)= 8.7 g MnF3

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