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the combustion of propane may be described by the chemical equation, 

C3H8(g) + 5O2(g) --> 3CO2(g) + 4H2O(g) 

how many grams of O2(g) are needed to completely burn 40.4 g of C3H8(g)?



User MSingh
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1 Answer

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\\u_(C_3H_8)=(40.4)/(\mu_(C_3H_8))=

=(40.4)/(3A_C+8A_H)=(40.4)/(44)=

=(9)/(10)=0.9mol

1mol\ C_3H_8...5\ mol\ O_2 \\ 0.9mol\ C_3H_8...x\ mol\ O_2

x=(0.9*5)/(1)=4.5mol\ O_2

m=\mu_(O_2)*x=2*A_O*4.5=9*8=72g\ O_2
User MWid
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