Question

Calculate the volume of oxygen gas produced under standard conditions when 6.8g of hydrogen peroxide is decomposed. 2H2O2(l) → 2H2O(l) + O2(g)​

Answer 1

no. of moles of H2O2 = 6.8/34 = 0.2

now, 2 moles of H2O2 produces 1 mole O2

therefore, 0.2 moles H2O2 will produce 0.1 mole of O2

now at stp 1 mole occupies a volume of 22.4L

therefore, 0.1 mole will occupy a volume of 2.24L