Finding percentage of the gas in a binary gaseous mixture of the given density - Chemistry Stack Exchange most recent 30 from chemistry.stackexchange.com 2019-09-22T05:44:25Z https://chemistry.stackexchange.com/feeds/question/108520 https://creativecommons.org/licenses/by-sa/4.0/rdf https://chemistry.stackexchange.com/q/108520 1 Finding percentage of the gas in a binary gaseous mixture of the given density Kai https://chemistry.stackexchange.com/users/73822 2019-01-25T05:57:04Z 2019-01-25T15:17:11Z <blockquote> <p>At STP, the density of a gas in a vessel is <span class="math-container">$0.9002$</span>. If the gas is a mixture of argon and helium, what percentage of the gas is argon?</p> </blockquote> <p>I am stuck on this. From what I can gather, the only influencing characters would be moles and grams. This idea is based of off of STP and given constants. </p> <p>I have a couple equations written down but I can’t seem to the flow going.</p> <p>R, T, P are all known along with acccording densities. I also want to say <span class="math-container">$22.4$</span> is also a known at <span class="math-container">$V_\mathrm{tot}$</span> and therefore <span class="math-container">$n_\mathrm{tot} = 1$</span>.</p> <p><span class="math-container">$$n(\ce{Ar}) + n(\ce{He}) = 1$$</span></p> <p><span class="math-container">$$\frac{n(\ce{Ar})}{n(\ce{Ar}) + n(\ce{He})}d(\ce{Ar}) + \frac{n(\ce{He})}{n(\ce{Ar}) + n(\ce{He})}d(\ce{He}) = d_\mathrm{tot}$$</span></p> <p><span class="math-container">$$d = PM/RT$$</span></p> https://chemistry.stackexchange.com/questions/108520/-/108537#108537 1 Answer by Buck Thorn for Finding percentage of the gas in a binary gaseous mixture of the given density Buck Thorn https://chemistry.stackexchange.com/users/69830 2019-01-25T10:11:52Z 2019-01-25T15:14:16Z <p>I'd start the solution with this equation:</p> <p><span class="math-container">$n_{Ar}d_{Ar}/(n_{Ar} + n_{He}) + n_{He}d_{He}/(n_{Ar} + n_{He}) = d_{avg}$</span></p> <p>where I emphasize that the density of the gas is a <em>molar average</em> value, and rewrite it as </p> <p><span class="math-container">$\chi_{Ar} d_{Ar} + (1-\chi_{Ar})d_{He} = d_{avg}$</span></p> <p>where <span class="math-container">$\chi_{Ar} = n_{Ar}/n$</span> is the mole fraction of argon in the gas mixture and the <span class="math-container">$d_{i}$</span> are densities computed assuming all of the gas corresponds to He or Ar, for instance</p> <p><span class="math-container">$d_{Ar} = M_{Ar}n/V = M_{Ar}/V_{m}$</span></p> <p>where <span class="math-container">$V_{m}$</span> is the molar volume at STP (22.414 <span class="math-container">$m^3/kg mol$</span>).</p> <p>It follows that </p> <p><span class="math-container">$d_{avg} = \chi_{Ar}M_{Ar}/V_{m} + (1-\chi_{Ar})M_{He}/V_{m} = (\chi_{Ar}M_{Ar} + (1-\chi_{Ar})M_{He})/V_{m} = M_{avg}/V_{m}$</span></p> <p>which can be solved for <span class="math-container">$\chi_{Ar}$</span>:</p> <p><span class="math-container">$\chi_{Ar} = ((d_{avg}V_{m})-M_{He})/(M_{Ar}-M_{He})$</span></p> <p>Finally, the molar percentage argon is computed as <span class="math-container">$f_{Ar}=100\chi_{Ar}$</span>.</p> <p>For your particular problem I get <span class="math-container">$f_{Ar}$</span>= 45.00%.</p> https://chemistry.stackexchange.com/questions/108520/-/108539#108539 3 Answer by William R. Ebenezer for Finding percentage of the gas in a binary gaseous mixture of the given density William R. Ebenezer https://chemistry.stackexchange.com/users/69977 2019-01-25T10:13:41Z 2019-01-25T15:17:11Z <p>Well, I can see all the relations you require knowledge of in the question itself!</p> <p>Calculate the effective molecular mass from </p> <p><span class="math-container">$$M = \frac{dRT}{P}$$</span> </p> <p><span class="math-container">$R$</span> is known, <span class="math-container">$d$</span> given. <span class="math-container">$P$</span> and <span class="math-container">$T$</span> are available from the fact that it is at STP.</p> <p><span class="math-container">$M$</span> comes out to be <span class="math-container">${20.176 u}$</span>.</p> <p>Effective molar mass is easily calculated below:</p> <p><span class="math-container">$$M_\mathrm{eq} = M_1x_1 + M_2x_2$$</span></p> <p>where <span class="math-container">$x_i$</span> is the mole fraction of each gas. You can replace <span class="math-container">$x_2$</span> by <span class="math-container">$1-x_1$</span>.</p> <p>Plugging in the values and solving for <span class="math-container">$x_1$</span>, argon mole fraction comes out to be <span class="math-container">$0.4493$</span>.</p> <p>I believe you can now calculate the demanded percentage, be it by mass or by moles.</p>