Question

Part A 11 45.0 g of gray iron powder reacts with 24.9 g of red phosphorus powder, what is the mass of the iron phosphide product? Express your answer with the appropriate units. HA 7 OF ? Value Units Submit Request Answer

Answer 1

Write the reaction between iron and phosphorus as follows: 3Fe+P — FezP 3 moles of Fe reacts with 1 mole of P and forms 1 mole of Fe P. mass of iron (Fe) = 45.0 g molar mass of Fe = 55.85 g/mol Calculate the moles of iron as follows: moles of Fe- mass of iron (Fe) 45.03 45.09 = 0.806 mol molar mass of Fe 55.85 g/mol ----------------------- mass of phosphorus (P) = 24.9 g molar mass of P = 30.97 g/mol Calculate the moles of phosphorus as follows: mass of phosphorus(P) 24.9g moles of P = -= 0.804 mol molar mass of P 30.97 g/mol -------------------- -- From the reaction the stoichiometric relation between Fe and Pis moles of Fe moles of P 3 Substitute the calculated moles in the above relation, the moles with least is the limitig reagent and the moles with highest is the excess reagent. 0.806 mol Fe 0.804 mol P - - - - - - - - - - - - - - 0.269 mol Fe < 0.804 mol P So the limiting reactant is Fe. limiting reactant is used to determine the amount of prouct produced From the reaction the stoichiometric relation between Fe and Fez Pis moles of Fe moles of FP moles of Fe, P=0.806 mol = 0.269 mol - - - - - - - - - - - - - - - - - - - - - - - molar mass of Fe, P = 198.5g/mol Calculate the mass of Fe, P as follows: mass of Fe, P= moles of Fe, Pxmolar mass of Fe,P = 0.269 mol 198.5g/mol = 53.4 g Therefore, the required solution is 53.4g|