Sheet 2 ere are two sources of error in this experiment, which fortuitously almo

Question

Sheet 2 ere are two sources of error in this experiment, which fortuitously almost cancel one another. Circle the correct answer in the following statements. (a) Initially the flask that you weigh is filled with air. At the end of the reaction it is filled with oxygen. Because oxygen is denser than air, at the end of the reaction the flask is (heavier, lighter) than it would be if filled with air. (b) If the oxygen that is evolved is saturated with water vapor, some water will be carried over as well as evolved oxygen. At the end of the reaction this effect would cause the flask to be (heavier, lighter) than if no water were carried over. Calculate the molarity of a solution that is 3.0% hydrogen peroxide by weight (3 g 11,02/100 g solution). assuming that the density of the solution is 1.00 g/mL. For one of your trials, calculate the hydrogen peroxide that you put in the flask, the moles of oxygen evolved, a moles of nd the ratio mol O,/mol H,02.s culate for the ratio consistent with the stoichiometry of the decomposition reaction? H2O2(aq) H2O(l) + O2(g)

Explanation / Answer

The balanced chemical equation for the decomposition reaction is

H2O2 (aq) -------> H2O (l) + ½ O2 (g)

We have a 3.0% H2O2 solution, i.e, 100 g solution contains 3.0 g H2O2.

Molar mass of H2O2 = (2*1.008 + 2*15.9994) g/mol = 34.0148 g/mol.

Moles of H2O2 corresponding to 3.0 g H2O2 = (3.0 g)/(34.0148 g/mol) = 0.088196 mole 0.0882 mole.

The density of the solution is 1.00 g/mL; therefore, the volume of the solution is (100 g)/(1.00 g/mL) = 100 mL = (100 mL)*(1 L/1000 mL) = 0.1 L.

Molar concentration of 3.0% H2O2 solution = (moles of H2O2)/(volume of solution in L) = (0.0882 mole)/(0.1 L) = 0.882 mol/L = 0.882 M (ans).

I need to know the experimental data to answer the remaining questions.

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