Please help me fill in the conversion factors and starting amount Consider the reaction of solid PbO and gaseous NH; to form solid Pb, nitrogengas, and water vapor.The balanced chemical equation is 3 PbO(s) + 2 NH3(g) → 3 Pb(s) + N2(g) + 3 H2O(g).How many moles of NH; are required to produce 185.6 grams of Pb based on thebalanced chemical equation?185.6g Pb2mol NH3207.2motP5X0.5972mol NH3%3D1motP53mol PbSTARTING AMOUNT185.6 g Pb2 mol NH3207.2 mol Pb х= 0.5972 mol NH31 mol Pb3 mol PbADD FACTORANSWERRESET*( )2207.210.59721.3440.895810.17185.617.03123.70.7443g Pbg NH3/mol NH3mol NH3g Pb/mol Pbmol Pbg NH3

The balanced chemical reaction of solid PbO and gaseous NH3 to formsolid Pb, nitrogen gas and water vapor is as follows,3PbO(s)+2NH3(g)→3Pb(s)+N2(g)+3H2O(g)amount of Pb produced in the reaction =185.6 gramwe are asked to calculate the number of moles of NH3 required to produce185.6 gram of PbThe balanced chemical reaction of solid PbO and gaseous NH3 to formsolid Pb, nitrogen gas and water vapor is as follows,3PbO(s)+2NH3(g)→3Pb(s)+N2(g)+3H2O(g)number of moles of Pb =given mass of Pbmolar mass of Pbnumber of moles of Pb =185.6 g207.2 g/mol =0.8958 molesnow, from the above balanced chemical reaction,since, 3 mole of Pb was produced from =2 mole of NH3so, 1 mole of Pb was produced from =23 mole of NH3thus, 0.896 mole of Pb was produced from =0.8958×23 mole of NH30.896 mole of Pb was produced from =0.5972 moles of NH3Therefore, the number of moles of NH3 required would be, 0.5972 moles. 185.6 g Pb × (1 mol Pb207.2 gPb) × (2 mol NH33 mol Pb) = 0.5972 mol NH3