First, we need to determine the moles of O2 produced from the decomposition of hydrogen peroxide. From the balanced chemical equation, we can see that for every 2 moles of H2O2, 1 mole of O2 is produced.
2H2O2 -> 2H2O + O2
Therefore, if 10.0 moles of H2O2 decompose, we can use the mole ratio to determine the moles of O2 produced:
10.0 moles H2O2 × (1 mole O2 / 2 moles H2O2) = 5.0 moles O2
Next, we will find the mass of oxygen produced using the molar mass of oxygen, O2 (molar mass of O = 16.00 g/mol; molar mass of O2 = 2 × 16.00 g/mol = 32.00 g/mol):
5.0 moles O2 × 32.00 g/mol = 160.0 g O2
Now, we can use the density of oxygen to find the volume of oxygen produced:
Density = Mass / Volume
1.429 g/L = 160.0 g / Volume
Volume = 160.0 g / 1.429 g/L = 111.9 L
Therefore, if 10.0 mol of H2O2 decomposes, it will produce approximately 111.9 L of oxygen gas.
Oxygen gas and water are produced by the decomposition of hydrogen peroxide. If 10.0 mol of H2O2 decomposes, what volume of oxygen will be produced? Assume the density of oxygen is 1.429g/L. 2H2O2= 2H2O+O2
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