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How many grams of oxygen is required for complete combustion of 29g of butane as per the equation

C4H10+ 4.5O2=2CO2+5H2O


Atomic weight of C4H10  = 58g
no. of mole of C4H10 = 29/58 = 1/2 mole
for 1 mole of butane combustion 9/2 mole of oxygen is required
for 1/2 mole of butane combustion 9/4 mole of oxygen is required
weight of oxygen = 9*16/4 = 36 gm.
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