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33317.03 g NH1.42755 mol NH1 mol NH= 24.311 g NH3% yield = actual yield100theoretical yield= 15.0 g10024.311 g= 61.700 = 61.7%2resent = 2221 mol H10.0 g H2.016 g H= 4.9602b) Moles of Hinitially p3 mol HMoles of N2initially present = 21 mol N220.0 g N28.02 g N= 0.71378 mol N22Moles of H2rered tooduce 15.0 g NHquipr3= 32315.0 g NH17.03 g33H3 mol HNH2 mol NH= 1.3212 mol H21 mol N323331 mol NH1 mol NMoles of N2required to produce 15.0 g NH3= 15.0 g NH17.03 g NH2 mol NH= 0.4404 mol N2Moles of Hat equilibrium = initial moles – reacted moles = 4.9603 mol – 1.3212 mol = 3.6391 = 3.64 mol H2rium = initial moles – reacted moles = 0.71378 mol – 0= 0.27338 = 0.273 mol N22Moles of N2at equilib.4404 mol 4-36
4.110 Plan:Ferrous ion is Fe2+. Write a reaction to show the conversion of Fe to Fe2+. Convert the mass of Fe in a 125-g Fe in a 737-g sample. Use molar massf Fe to moles of Fe and use Avogadro’s number to convert moles of Fe to moles of ions. Solution:serving to the mass of to convert mass o=a) Fe oxidizes to Fe2+with a loss of 2 electrons. The H+in the acidic food is reduced to H2with a gain of 2 electrons. The balanced reaction is: Fe(s) + 2H+(aq) Fe2+(aq) + H2(g) O.N.: 0 +1 +2 0 b) Mass (g) of Fe in the jar of tomato sauce349 mg Fe10g737 g sauce125 g sauce1 mg= 0.288904 g Fe ons = 223221 mol Fe1 mol Fe6.022x10mol Fe1 molNumber of Fe2+iFeions0.288904 g Fe55.85 g Fe1Fe= 3.11509x1021= 3.1x1021Fe2+ionsper jar of sauce 4.111Plan:Write balanced equations for the conversion of CaCO3to CaO, the reaction of CaO with SO2,and the three reactions to obtain the overall reaction. Use the mass % of d the mass and then moles of sulfur present. Use the molar ratio in the overall ass of CaCO3required toct with this amount of sulfur, asming a 70% yield. combustion of sulfur to produce SO2. Add thesesulfur in the coal sample to finbalanced reaction to find the mreasuSolution:The reactions are: CaCO (s)CaO(3s)+ CO2(g) CaO(s)+ SO2(g)CaSO3(s) S(s) + O2(g) SO2(g)Overall reaction: CaCO3(s) + S(s) + O2(g) CaSO3(s) + CO2(g) Moles of sulfur = 3410 g0.33%1 mol S8.5x10 kg coal1 kg100%32.07 g S= 8746.492 mol S Mass (g) CaCO3= 31 mol CaCOl S33100.09 g CaCO100%1 mol S1 mol CaCO70%6= 1.3x106g CaCO3needed 4.112 Plan:8746.492 mo= 1.25062x10Convert the mass of glucose to moles and use the molar ratios from the balanced equation to find the and CO2. The amount of ethanol is cograms using its molar mass.