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

User Richie Rizal Amir
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1 Answer

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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
User Tagore Smith
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