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Question

owv2 Online teachin X takeAssignment/tak ty.do?locator assignment tor-assignment-tak Use the Refereaces to access imsportant values if needed for this question. For the following reaction, 24.8 grams of iron are allowed to react with irom(s) + hydrochloric acid(ag)iroaI chloride(aq)+ hydrogen(g) what is the maximum amount of iron(I) chloride that can be formed? L What is the FORMULA for the limiting reagent? 27.4 grams of hydrochloric acid grams What amount of the excess reagent remains after the reaction is complete? grams Submit Answer Retry Entire Group 9 more group attempts remaining

Explanation / Answer

Fe(s) + 2HCl(aq) --------> FeCl2(aq) + H2(g)
no of moles of Fe = W/G.M.Wt
                   = 24.8/56   = 0.443 moles
no of moles of HCl = W/G.M.Wt
                    = 27.4/36.5 = 0.75 moles
1 mole of Fe react with 2 mole of HCl
0.443 moles of Fe react with = 2*0.443/1 = 0.886 moles of HCl
HCl is limiting reagent
2 moles of HCl react with excess of HCl to gives 1 moles of FeCl2
0.75 moles of HCl react with excess of HCl to gives = 1*0.75/2 = 0.375 moles of FeCl2
mass of FeCl2 = no of moles * gram molar mass
               = 0.375*126.75   = 47.53g of FeCl2
2 moles of HCl react with 1 mole of Fe
0.75 moles of HCl react with = 1*0.75/2 = 0.375 moles of Fe
Fe is excess reacgent
The no of moles of excess reagent remains after complete the reaction = 0.443-0.375   = 0.068 moles
The amount of excess reagent remains after complete the reaction = no of moles *gram molar mass
                                                                  = 0.068*56 = 3.808g of Fe

2.
2CO(g) + O2(g) --------> 2CO2(g)
no of moles of CO   = W/G.M.Wt
                    = 9.78/28   = 0.35 moles
no of moles of O2    = W/G.M.Wt
                     = 9.81/32   = 0.31 moles
1 mole of O2 react with 2 moles of CO
0.31 moles of O2 react with = 2*0.31/1 = 0.62 moles of CO is required
Co is limiting reagent
2 moles of CO react with excess O2 to gives 2moles of CO2
0.35 moles of CO react with excess O2 to gives = 2*0.35/2 = 0.35 moles of CO2
mass of CO2 = no ofmoles * gram molar mass
             =0.35*44 = 15.4g of CO2
the theoretical yield of CO2 = 15.4g >>>answer
2 moles of CO react with 1 moles of O2
0.35 moles of CO react with = 1*0.35/2   = 0.175 moles of O2
O2 is excess reagent
The no of moles of excess reagent remains after complete the reaction = 0.35-0.175   = 0.175 moles
The amount of excess reagent remains after complete the reaction = no of moles *gram molar mass
                                                                 = 0.175*32 = 5.6g

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