{"id":229079,"date":"2025-06-08T01:57:10","date_gmt":"2025-06-08T01:57:10","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=229079"},"modified":"2025-06-08T01:57:13","modified_gmt":"2025-06-08T01:57:13","slug":"analysis-of-an-aluminum-zinc-alloy-trial-mass-gelatin-capsule-2134-3218-1542-406-723-mass-of-alloy-sample-plus-capsule-mass-empty-beaker-mass-of-beaker-plus-displaced-water-atmospheric-pressure-temp-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/08\/analysis-of-an-aluminum-zinc-alloy-trial-mass-gelatin-capsule-2134-3218-1542-406-723-mass-of-alloy-sample-plus-capsule-mass-empty-beaker-mass-of-beaker-plus-displaced-water-atmospheric-pressure-temp-2\/","title":{"rendered":"Analysis of an Aluminum-Zinc Alloy Trial [ Mass gelatin capsule 2134 3218 1542 406 723 Mass of alloy sample plus capsule Mass empty beaker Mass of beaker plus displaced water Atmospheric pressure Temperature Calculations Mass alloy sample Mass of displaced water Volume of displaced water (density of water: 1.00 g\/mL) Volume of H2 gas Temperature of H2 Vapor pressure of water at room temperature"},"content":{"rendered":"\n<pre id=\"preorder-ask-header-text\" class=\"wp-block-preformatted\">Analysis of an Aluminum-Zinc Alloy Trial [ Mass gelatin capsule 2134 3218 1542 406 723 Mass of alloy sample plus capsule Mass empty beaker Mass of beaker plus displaced water Atmospheric pressure Temperature Calculations Mass alloy sample Mass of displaced water Volume of displaced water (density of water: 1.00 g\/mL) Volume of H2 gas Temperature of H2 Vapor pressure of water at room temperature: 17.5 mmHg (Appendix I) n Hg Pressure of dry H2 gas Hg: AlTi MOK > Moles H2 in sample, \"it; Moles H2 per gram of sample. Nil; mokvg Al (read from graph) Unknown<\/pre>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Given Data:<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Mass of gelatin capsule:<\/strong> (2134, 3218, 1542, 406, 723) \u2013 likely trial masses; we\u2019ll select one value as an example.<\/li>\n\n\n\n<li><strong>Mass of alloy sample plus capsule:<\/strong> (value not explicitly separated)<\/li>\n\n\n\n<li><strong>Mass of empty beaker:<\/strong> not specified<\/li>\n\n\n\n<li><strong>Mass of beaker plus displaced water:<\/strong> not specified<\/li>\n\n\n\n<li><strong>Atmospheric pressure:<\/strong> not specified; assume <strong>760 mmHg<\/strong> (standard)<\/li>\n\n\n\n<li><strong>Temperature:<\/strong> not specified; assume <strong>25\u00b0C<\/strong><\/li>\n\n\n\n<li><strong>Vapor pressure of water at 25\u00b0C:<\/strong> 17.5 mmHg<\/li>\n\n\n\n<li><strong>Density of water:<\/strong> 1.00 g\/mL<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">Let\u2019s <strong>assume<\/strong> and reconstruct plausible values to perform an example calculation.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Example Data (Assumed from structure):<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Mass of alloy + capsule:<\/strong> 3.218 g<\/li>\n\n\n\n<li><strong>Mass of capsule:<\/strong> 2.134 g<br>\u2192 <strong>Mass of alloy sample:<\/strong> 3.218 g \u2013 2.134 g = <strong>1.084 g<\/strong><\/li>\n\n\n\n<li><strong>Mass of displaced water:<\/strong> 154.2 g<br>\u2192 <strong>Volume of water displaced = 154.2 mL = volume of H\u2082 gas collected<\/strong><\/li>\n\n\n\n<li><strong>Temperature of gas = 25\u00b0C = 298 K<\/strong><\/li>\n\n\n\n<li><strong>Atmospheric pressure:<\/strong> 760 mmHg<\/li>\n\n\n\n<li><strong>Vapor pressure of water:<\/strong> 17.5 mmHg<br>\u2192 <strong>Pressure of dry H\u2082 gas:<\/strong> 760 \u2013 17.5 = <strong>742.5 mmHg<\/strong> = <strong>0.977 atm<\/strong><\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>1. Moles of H\u2082 Gas (Ideal Gas Law):<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Use:<br>PV=nRTPV = nRTPV=nRT<br>n=PVRTn = \\frac{PV}{RT}n=RTPV\u200b<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>P:<\/strong> 0.977 atm<\/li>\n\n\n\n<li><strong>V:<\/strong> 154.2 mL = 0.1542 L<\/li>\n\n\n\n<li><strong>R:<\/strong> 0.0821 L\u00b7atm\/mol\u00b7K<\/li>\n\n\n\n<li><strong>T:<\/strong> 298 K<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">n=(0.977)(0.1542)(0.0821)(298)\u22480.150624.486\u22480.00615&nbsp;mol&nbsp;H2n = \\frac{(0.977)(0.1542)}{(0.0821)(298)} \\approx \\frac{0.1506}{24.486} \\approx 0.00615 \\text{ mol H}_2n=(0.0821)(298)(0.977)(0.1542)\u200b\u224824.4860.1506\u200b\u22480.00615&nbsp;mol&nbsp;H2\u200b<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>2. Moles of H\u2082 per gram of sample:<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">0.00615&nbsp;mol1.084&nbsp;g=0.00567&nbsp;mol\/g\\frac{0.00615 \\text{ mol}}{1.084 \\text{ g}} = 0.00567 \\text{ mol\/g}1.084&nbsp;g0.00615&nbsp;mol\u200b=0.00567&nbsp;mol\/g<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>3. Determine % Aluminum Using Calibration Graph:<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Let\u2019s assume a calibration graph indicates <strong>moles H\u2082\/g<\/strong> for pure Al is 0.037 and pure Zn is 0 (Zn does not produce H\u2082 with acid in this context). Then: %Al=0.005670.037\u00d7100\u224815.3%\\% \\text{Al} = \\frac{0.00567}{0.037} \\times 100 \\approx 15.3\\%%Al=0.0370.00567\u200b\u00d7100\u224815.3%<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Conclusion<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">This experiment determines the composition of an aluminum-zinc alloy by analyzing the volume of hydrogen gas produced when the alloy reacts with hydrochloric acid. Aluminum reacts with HCl to release hydrogen gas: 2Al+6HCl\u21922AlCl3+3H22Al + 6HCl \\rightarrow 2AlCl_3 + 3H_22Al+6HCl\u21922AlCl3\u200b+3H2\u200b<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Zinc may also react with HCl, but in some versions of this experiment, it is assumed that aluminum is the primary H\u2082 source due to conditions or prior calibration. The volume of hydrogen gas collected over water reflects the aluminum content. However, because water vapor also exerts pressure, we must correct the pressure of the collected gas by subtracting the vapor pressure of water at the experiment temperature (17.5 mmHg at 25\u00b0C).<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Using the Ideal Gas Law, the number of moles of hydrogen gas produced is calculated. This value, divided by the mass of the alloy sample, gives the number of moles of hydrogen per gram of sample. This value is then compared to standard values obtained from a calibration graph\u2014pure aluminum produces a known maximum amount of hydrogen, while zinc does not. The proportion of hydrogen gas produced per gram directly correlates with the percentage of aluminum in the sample.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">In our calculation, the sample produced 0.00615 mol H\u2082 from 1.084 g alloy, equating to 0.00567 mol\/g. Comparing this to the 0.037 mol\/g standard for pure aluminum, we estimate that the alloy is approximately 15.3% aluminum by mass. The rest of the alloy is presumably zinc, completing the analysis.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner7-312.jpeg\" alt=\"\" class=\"wp-image-229080\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Analysis of an Aluminum-Zinc Alloy Trial [ Mass gelatin capsule 2134 3218 1542 406 723 Mass of alloy sample plus capsule Mass empty beaker Mass of beaker plus displaced water Atmospheric pressure Temperature Calculations Mass alloy sample Mass of displaced water Volume of displaced water (density of water: 1.00 g\/mL) Volume of H2 gas Temperature [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-229079","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/229079","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=229079"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/229079\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=229079"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=229079"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=229079"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}