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Hydrogen peroxide decomposes to give water and oxygen gas according to the equation below. 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? 2 H2O2(l) → 2 H2O(l) + O2(g)

Respuesta :

First you need to use stoichiometry to find how many moles of oxygen is produced from the reaction.  To do this you need to multiply the number of moles of hydrogen peroxide by the molar ratio which is 1/2 since 1 mole of O₂ is produced from 1 mole of H₂O₂.  3mol H₂O₂x(1mol O₂/2mol H₂O₂)=1.5mol O₂
you you need to you the ideal gas law (PV=nRT) to find the volume of gas produced. V=nRT/P
n=1.5mol
R=0.08206atmL/molK
T=23°C (turn that into 296K)
P=1atm
V=(1.5molx0.08206atmL/molK)/1atm
V=0.1231L

I hope this helps.

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