What is the pressure generated in a 10 liter container at 300K when
68 gram of H2O2 decomposes?
please I only have 30 minutes left

                     Pressure  =  2.45 atm

Explanation:

                    The balance chemical equation for the decomposition of hydrogen peroxide is as follow;

                                 2 H₂O₂ ₍aq₎ → 2 H₂O ₍l₎ + O₂ ₍g₎

Step 1: Calculate Moles of H₂O₂ as;

                     Moles  =  Mass / M.Mass

                     Moles  =  68.0 g / 34.01 g/mol

                     Moles  =  1.99 moles

Step 2: Calculate Moles of O₂ gas produced:

According to equation,

2 moles of H₂O₂ produced  =  1 mole of O₂

So,

1.99 moles of H₂O₂ will produce  =  X moles of O₂

Solving for X,

                     X  =  1.99 mol × 1 mol / 2 mol

                      X =  0.995 moles of O₂

Step 3: Calculate pressure exerted by O₂ gas as;

According to ideal gas equaton,

                                     P V  =  n R T

Or,

                                     P  =  n R T / V

Putting values,

                   P  =  0.995 mol × 0.08205 L.atm.K⁻¹.mol⁻¹ × 300 K / 10 L

                   P  =  2.45 atm

20 drops hope this

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