

Stoichiometry - Engineering
Q1: Pure oxygen is mixed with air to produce an enriched air containing 50 volume % of oxygen. The ratio of moles of air to oxygen used isA 1.72
B 0.58
C 0.5
D 0.2
ANS:B - 0.58 To solve this problem, let's denote the volume of pure oxygen as 𝑉oxygenVoxygen and the volume of air as 𝑉airVair. Since the enriched air contains 50 volume % of oxygen, it means that the volume of oxygen is half the total volume of the enriched air: 𝑉oxygen=0.5×(𝑉oxygen+𝑉air)Voxygen=0.5×(Voxygen+Vair) Given that pure oxygen is mixed with air, we know that the volume of oxygen used is equal to the volume of pure oxygen used: 𝑉oxygen=𝑉pure oxygenVoxygen=Vpure oxygen Let's denote the total volume of the mixture as 𝑉mixtureVmixture, then we have: 𝑉mixture=𝑉oxygen+𝑉airVmixture=Voxygen+Vair From the first equation, we can express 𝑉oxygenVoxygen in terms of 𝑉airVair: 𝑉oxygen=0.5×(𝑉oxygen+𝑉air)Voxygen=0.5×(Voxygen+Vair) 𝑉oxygen=0.5×𝑉mixtureVoxygen=0.5×Vmixture Since 𝑉oxygen=𝑉pure oxygenVoxygen=Vpure oxygen, we have: 𝑉pure oxygen=0.5×𝑉mixtureVpure oxygen=0.5×Vmixture Now, let's consider the mole ratios: moles of oxygen=volume of pure oxygenmolar volume of oxygen at STPmoles of oxygen=molar volume of oxygen at STPvolume of pure oxygen moles of air=volume of airmolar volume of air at STPmoles of air=molar volume of air at STPvolume of air The ratio of moles of air to oxygen used is: moles of airmoles of oxygen=𝑉air𝑉molar volume of air0.5×𝑉mixture𝑉molar volume of oxygenmoles of oxygenmoles of air=Vmolar volume of oxygen0.5×VmixtureVmolar volume of airVair At standard temperature and pressure (STP), the molar volume of oxygen is approximately 22.414 L/mol, and the molar volume of air (which is essentially the same as the molar volume of nitrogen, since air is mostly nitrogen) is also approximately 22.414 L/mol. Substituting these values: 𝑉air22.4140.5×𝑉mixture22.414=𝑉air0.5×𝑉mixture22.4140.5×Vmixture22.414Vair=0.5×VmixtureVair Now, using the fact that 𝑉mixture=𝑉oxygen+𝑉airVmixture=Voxygen+Vair, we can rewrite 𝑉airVair as 𝑉mixture−𝑉oxygenVmixture−Voxygen: 𝑉mixture−𝑉oxygen0.5×𝑉mixture0.5×VmixtureVmixture−Voxygen =𝑉mixture0.5×𝑉mixture−𝑉oxygen0.5×𝑉mixture=0.5×VmixtureVmixture−0.5×VmixtureVoxygen =2−1=2−1 =1=1 So, the ratio of moles of air to oxygen used is 1. Therefore, none of the provided options match the correct ratio. |


For help Students Orientation
Mcqs Questions
One stop destination for examination, preparation, recruitment, and more. Specially designed online test to solve all your preparation worries. Go wherever you want to and practice whenever you want, using the online test platform.