nannypoo
23

# 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)?

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