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8 August, 14:11

How many grams of O2 (g) are needed to completely burn 39.1 g of C3H8 (g) ?

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  1. 8 August, 17:52
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    The formula is

    CxHy + (x+y/4) O2=xCO2+yH2O

    the molecular weight of C3H8 is 44 gm so according the formula (5*32) gm O2 are needed to completely burn 44gm C3H8 so for 39.1 gm C3H8 we needed (39.1*32*5) / (44) = 142.181 gm
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