Decomposition of Hydrogen Peroxide Oxygen was produced by the decomposition of h

Question

Decomposition of Hydrogen Peroxide Oxygen was produced by the decomposition of hydrogen peroxide (H_2O_2) according to the 2H_2O(aq) rightarrow O_2(g) + 2H_2O(I) A 5.0-ml aliquot of hydrogen peroxide solution produced a total of 40.2 ml of O_2 gas wit temperature at 35.0 degree C and O_2 pressure at 718 mmHg. Calculate the moles of oxygen evolved. (Use the Ideal gas law and R = 0.08206 L.atm/Based on the mole of O_2 collected, calculate the molar concentration of the 5-mL (H_2O_2) used. 2H_2O_2(aq) rightarrow O_2(g) + 2H_2O(I)

Explanation / Answer

Ans 1 a)

According to the ideal gas law

pV = nRT

where p is pressure , V is volume , n is number of moles of gas , R is universal gas constant and T is temperature in kelvin

Putting all the terms in the formula above

718 mm Hg= 0.94474 atm

0.94474 x 0.0402 = n x 0.08206 x 308

n = 0.0015 moles

Number of moles of oxygen gas = 0.0015

b)

2 moles of hydrogen peroxide produces one mole of oxygen on decomposition

0.0015 moles of oxygen gas will be produced by 0.0015 x 2 = 0.003 moles of hydrogen peroxide.

Molarity = number of moles/ volume of solution in litre

M = 0.003 / 0.005

M = 0.6

So the molar concentration of hydrogen peroxide will be 0.6 M

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