"1mol C_6H_{10}" reacts with "8.5mol O_2"
Moles of "C_6H_{10}=\\frac{35}{82}=0.426mol"
Moles of "O_2=\\frac{45}{32}=1.406mol"
As "0.165mol C_6H_{10}" is required to completely react with "1.406mol O_2"
So "O_2" is a limiting reagent
Moles of "CO_2" formed when "1.406mol O_2" reacted"=1.406\\times\\frac{12}{17}=0.992mol"
Hence mass of "CO_2" formed"=0.992\\times44=43.648g"
Moles of excess "C_6H_{10}" left over after reaction"=0.426-0.165=0.261mol"
If "35g CO_2" is formed then percent yield"\\frac{35}{43.648}\\times100=80.18"
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