[tex]98.56 dm^3[/tex] of oxygen at STP would be required to react completely with 38.8g of propane.
Given that :
molar mass of propane = 44 g/mol
mass of propane = 38.8 g
∴ Moles present in 38.8 g of propane = [tex]\frac{38.8}{44}[/tex] = 0.88 mole
applying rule of balanced equations
1 mole of propane = 5 moles of oxygen
0.88 mole of propane = 5 * 0.88 = 4.4 moles of oxygen
Note : volume of 1 mole of oxygen at STP = [tex]22.4 dm^3[/tex]
∴Total volume of oxygen required at STP = 22.4 * 4.4 = [tex]98.56 dm^3[/tex]
Hence we can conclude that the volume of oxygen at STP required to react completely [tex]98.56 dm^3[/tex]
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