What volume of oxygen can be obtained by the addition of excess water to 1.0g of sodium peroxide?

Question

HIX Tutor · Accepted Answer

undefined To calculate the volume of oxygen obtained:
Step 1: Calculate the number of moles in 1.0g of sodium peroxide.
Step 2: Write and balance the chemical equation for the reaction of sodium peroxide with water to produce oxygen.
Step 3: Determine the stoichiometry where 1 mole of sodium peroxide produces 1 mole of oxygen.
Step 4: Use the ideal gas law (PV = nRT) to calculate the volume of oxygen by knowing the number of moles of oxygen at standard temperature and pressure (STP).
Step 5: Plug in the values into the formula V = nRT/P to find the volume of oxygen.

Noah Aldrich · Answer

#"0.15 L " -># at STP

First things first: without knowledge of the pressure and temperature parameters, it is impossible to determine the amount of oxygen produced by the reaction. SInce no mention of this was made, I'll assume that the reaction takes place at STP - Standard Temperature and Pressure, which implies a pressure of #"100 kPa"# and a temperature of #0^@"C"#. Write the balanced chemical equation for this reaction with your current focus. Sodium peroxide, #"Na"_2"O"_2#, will react with water to form sodium hydroxide, #"NaOH"#, and hydrogen peroxide, #"H"_2"O"_2#. #"Na"_2"O"_text(2(s]) + 2"H"_2"O"_text((l]) -> 2"NaOH"_text((aq]) + "H"_2"O"_text(2(aq])# The breakdown of the hydrogen peroxide will be the source of the oxygen gas. #2"H"_2"O"_text(2(aq]) -> 2"H"_2"O"_text((l]) + "O"_text(2(g]) uarr# This formula can be expressed as #"H"_2"O"_text(2(aq]) -> "H"_2"O"_text((l]) + 1/2"O"_text(2(g]) uarr# To determine the overall reaction, combine these two equations. #"Na"_2"O"_text(2(s]) + color(red)(cancel(color(black)(2)))"H"_2"O"_text((l]) -> 2"NaOH"_text((aq]) + color(red)(cancel(color(black)("H"_2"O"_text((l])))) + 1/2"O"_text(2(g]) uarr# which results in #"Na"_2"O"_text(2(s]) + "H"_2"O"_text((l]) -> 2"NaOH"_text((aq]) + 1/2"O"_text(2(g]) uarr# Notice that the reaction produces #1/2# moles of oxygen gas for every mole of sodium peroxide. Utilizing the compound's molar mass, figure out how many moles of sodium peroxide you have. #1.0color(red)(cancel(color(black)("g"))) * ("1 mole Na"_2"O"_2)/(77.978color(red)(cancel(color(black)("g")))) = "0.0128 moles Na"_2"O"_2# The outcome of the reaction will be #0.0128color(red)(cancel(color(black)("moles Na"_2"O"_2))) * ("1/2 moles O"_2)/(1color(red)(cancel(color(black)("mole Na"_2"O"_2)))) = "0.00640 moles O"_2# At STP, one mole of any ideal gas occupies exactly #"22.7 L"# - this is known as the molar volume of a gas at STP. This many moles of oxygen gas would occupy in your situation #0.00640color(red)(cancel(color(black)("moles"))) * "22.7 L"/(1color(red)(cancel(color(black)("mole")))) = color(green)("0.15 L")#

Naomi Aronson · Answer

undefined 22.4 liters of oxygen can be obtained by the addition of excess water to 1.0g of sodium peroxide.