Calculate the mass of oxygen (in grams) required to completely burn 4.48 liters of propane.

univerkov | December 9, 2020 | education

1) Let’s write the reaction C3H8 + 5O2 -> 3CO2 + 4H2O
2) Find the number of moles of propane n (C3H8) = V (C3H8) / Vm n (C3H8) = 4.48 l / 22.4 l / mol = 0.2 mol
3) Let’s make a proportion of 1 – 0.2 mol 5 – x mol From where x = 1 mol Found a mole of oxygen
4) m (O2) = 1mol * 32g / mol = 32g
Answer: 32g

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