Question

How many dm3 of oxygen at STP would be required to react completely with 38.8g of propane? (3 marks)

Answer 1

$98.56 dm^3$ 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 = $\frac{38.8}{44}$ = 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 = $22.4 dm^3$

∴Total volume of oxygen required at STP = 22.4 * 4.4 = $98.56 dm^3$

Hence we can conclude that the volume of oxygen at STP required to react completely $98.56 dm^3$

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