Answered

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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)?

Sagot :

[tex]\nu_{C_3H_8}=\frac{40.4}{\mu_{C_3H_8}}=[/tex]
[tex]=\frac{40.4}{3A_C+8A_H}=\frac{40.4}{44}=[/tex]
[tex]=\frac{9}{10}=0.9mol[/tex]
[tex]1mol\ C_3H_8...5\ mol\ O_2 \\ 0.9mol\ C_3H_8...x\ mol\ O_2[/tex]
[tex]x=\frac{0.9*5}{1}=4.5mol\ O_2[/tex]
[tex]m=\mu_{O_2}*x=2*A_O*4.5=9*8=72g\ O_2[/tex]
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