Hydrogen peroxide decomposes to give water and oxygen gas according to the equation 2 H2O2(l) ? 2 H2O(l) + O2(g). If 3.0 moles of hydrogen peroxide decompose, what volume of oxygen gas is produced at a pressure of 1.0 atm and a temperature of 23 °C? 1.9 L of O2 2.8 L of O2 24 L of O2 36 L of O2
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According to the balanced chemical equation, 2 moles of hydrogen peroxide (H2O2) decompose to produce 1 mole of oxygen gas (O2). So, if 3.0 moles of H2O2 decompose, we would expect 1.5 moles of O2 to be produced (since 3.0 moles H2O2 * (1 mole O2 / 2 moles H2O2) Show more…
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The following reaction illustrates the decomposition of hydrogen peroxide (H2O2) to form water and oxygen gas: 2 H2O2 (l) → 2 H2O (l) + O2 (g). When this reaction is carried out, 0.34 L of O2 gas is collected over water at a total pressure of 1.09 atm. The temperature is 23°C. The vapour pressure of water at this temperature is 0.0278 atm. What was the mass (in g) of the hydrogen peroxide that decomposed?
Chareen G.
hydrogen peroxide (H2O2) decomposes to form water and oxygen gas. At STP, how many mLs of oxygen can be produced when 10.0 grams of hydrogen peroxide decomposes?
Adi S.
Hydrogen peroxide, $\mathrm{H}_{2} \mathrm{O}_{2},$ decomposes to form water and oxygen. \begin{equation}\begin{array}{l}{\text { a. How many liters of } \mathrm{O}_{2} \text { can be made }} \\ {\text { from } 342 \mathrm{g} \mathrm{H}_{2} \mathrm{O}_{2} \text { if the density of } \mathrm{O}_{2} \text { is }} \\ {1.428 \mathrm{g} / \mathrm{L} ?} \\ {\text { b. The density of } \mathrm{H}_{2} \mathrm{O}_{2} \text { is } 1.407 \mathrm{g} / \mathrm{mL}, \text { and }} \\ {\text { the density of } \mathrm{O}_{2} \text { is } 1.428 \mathrm{g} / \mathrm{L} . \text { How }} \\ {\text { many liters of } \mathrm{O}_{2} \text { can be made from }} \\ {55 \mathrm{mL} \mathrm{H}_{2} \mathrm{O}_{2} ?}\end{array}\end{equation}
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