What mass of O2 remains un-reacted when 2.00 g of Al reacts with 50.0 g of O2 to

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Explanation / Answer

moles of Al=2g/(26.98g/gmol)=7.4129×10^-2gmol moles of O2=50g/(32g/gmol)=1.5625gmol reaction is given by 4Al+3O2==>2Al2O3 it means 4 moles aluminum reacts with 3 moles of O2 to give 2 moles of Al2O3 first we have to calculate limiting reactant first for O2 3 moles of O2 give =2 moles of Al2O3 1 moles of O2 give =2/3 moles of Al2O3 1.5625 moles of O2 give =2/3 ×1.5625moles of Al2O3=1.0417gmol Al2O3 now for Al 4 moles of Al give =2 moles of Al2O3 1 moles of Al give =2/4 moles of Al2O3 7.4129×10^-2 moles of Al give =2/4 ×7.4129×10^-2moles of Al2O3=0.037065gmol Al2O3 so Al is limiting reactant now for O2 consumed 4 mol Al react with =3mol O2 1mol Al react with =3/4 mol O2 7.4129×10^-2mol Al react with =3/4×7.4129×10^-2 mol O2=0.05559mol O2 O2 remaining= 1.5625mol-0.05559mol=1.5069mol O2 mass of O2 remaining=1.5069×32=48.221g

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