{"id":195269,"date":"2025-02-28T05:42:51","date_gmt":"2025-02-28T05:42:51","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=195269"},"modified":"2025-02-28T05:42:54","modified_gmt":"2025-02-28T05:42:54","slug":"use-the-following-balanced-equation","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/28\/use-the-following-balanced-equation\/","title":{"rendered":"Use the following balanced equation"},"content":{"rendered":"\n

Use the following balanced equation,<\/p>\n\n\n\n

CaBr2(aq) + H2SO4(aq) ? CaSO4(s) + 2HBr(g)<\/p>\n\n\n\n

If 10.32 g of HBr are generated, what volume of 0.100 M H2SO4 is needed to produce this much HBr?<\/p>\n\n\n\n

lewis structure for SbF5?<\/p>\n\n\n\n

The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n

Let’s start with the first problem:<\/p>\n\n\n\n

Given balanced equation:<\/h3>\n\n\n\n

[
\\text{CaBr}_2(aq) + \\text{H}_2\\text{SO}_4(aq) \\rightarrow \\text{CaSO}_4(s) + 2\\text{HBr}(g)
]<\/p>\n\n\n\n

Step 1: Convert Grams of HBr to Moles<\/h3>\n\n\n\n

The molar mass of HBr is calculated as:
[
\\text{HBr} = 1.008 + 79.904 = 80.912 \\text{ g\/mol}
]<\/p>\n\n\n\n

Given:
[
10.32 \\text{ g HBr} \\times \\frac{1 \\text{ mol HBr}}{80.912 \\text{ g HBr}} = 0.1275 \\text{ mol HBr}
]<\/p>\n\n\n\n

Step 2: Determine Moles of H\u2082SO\u2084 Needed<\/h3>\n\n\n\n

From the balanced equation, 2 moles of HBr are produced per 1 mole of H\u2082SO\u2084<\/strong>, so:<\/p>\n\n\n\n

[
0.1275 \\text{ mol HBr} \\times \\frac{1 \\text{ mol H}_2\\text{SO}_4}{2 \\text{ mol HBr}} = 0.06375 \\text{ mol H}_2\\text{SO}_4
]<\/p>\n\n\n\n

Step 3: Calculate Volume of H\u2082SO\u2084 Solution<\/h3>\n\n\n\n

Given that the concentration of H\u2082SO\u2084<\/strong> is 0.100 M<\/strong>:<\/p>\n\n\n\n

[
\\text{Volume} = \\frac{\\text{moles of H}_2\\text{SO}_4}{\\text{Molarity}} = \\frac{0.06375}{0.100} = 0.6375 \\text{ L} = 637.5 \\text{ mL}
]<\/p>\n\n\n\n

Final Answer:<\/h3>\n\n\n\n

To produce 10.32 g of HBr<\/strong>, 637.5 mL of 0.100 M H\u2082SO\u2084<\/strong> is required.<\/p>\n\n\n\n

\n\n\n\n

2. Lewis Structure of SbF\u2085<\/h3>\n\n\n\n

To determine the Lewis structure<\/strong> for SbF\u2085<\/strong> (Antimony Pentafluoride):<\/p>\n\n\n\n

    \n
  1. Count Valence Electrons:<\/strong><\/li>\n<\/ol>\n\n\n\n
      \n
    • Sb (Antimony) is in Group 15<\/strong> \u2192 5 valence electrons<\/strong>.<\/li>\n\n\n\n
    • F (Fluorine) is in Group 17<\/strong> \u2192 7 valence electrons<\/strong>.<\/li>\n\n\n\n
    • Total for SbF\u2085<\/strong> = 5 + (5 \u00d7 7) = 40 valence electrons<\/strong>.<\/li>\n<\/ul>\n\n\n\n
        \n
      1. Draw the Skeleton Structure:<\/strong><\/li>\n<\/ol>\n\n\n\n
          \n
        • Place Sb<\/strong> in the center (less electronegative).<\/li>\n\n\n\n
        • Attach five<\/strong> fluorine atoms around it.<\/li>\n<\/ul>\n\n\n\n
            \n
          1. Distribute Electrons:<\/strong><\/li>\n<\/ol>\n\n\n\n
              \n
            • Each F<\/strong> needs 8 electrons<\/strong> (octet rule).<\/li>\n\n\n\n
            • All five fluorines are single bonded<\/strong> to Sb, using 10 bonding electrons (5 bonds \u00d7 2 electrons)<\/strong>.<\/li>\n\n\n\n
            • Remaining 30 electrons<\/strong> go to fluorines as lone pairs (6 electrons per F<\/strong>).<\/li>\n<\/ul>\n\n\n\n
                \n
              1. Check Formal Charges:<\/strong><\/li>\n<\/ol>\n\n\n\n
                  \n
                • Sb has five single bonds<\/strong> (0 lone pairs).<\/li>\n\n\n\n
                • Each F has three lone pairs<\/strong> and one bond.<\/li>\n\n\n\n
                • All atoms have formal charges of zero<\/strong>, making it stable.<\/li>\n<\/ul>\n\n\n\n

                  Since SbF\u2085<\/strong> follows the AX\u2085<\/strong> molecular geometry (VSEPR theory), it has a trigonal bipyramidal shape<\/strong>.<\/p>\n\n\n\n

                  I will generate the Lewis Structure image<\/strong> now.<\/p>\n\n\n\n

                  \"\"<\/figure>\n\n\n\n

                  Here is the Lewis structure diagram for SbF\u2085. You can see that the central Sb atom is bonded to five fluorine atoms in a trigonal bipyramidal arrangement, with each fluorine obeying the octet rule. Let me know if you need any further clarification!<\/p>\n","protected":false},"excerpt":{"rendered":"

                  Use the following balanced equation, CaBr2(aq) + H2SO4(aq) ? CaSO4(s) + 2HBr(g) If 10.32 g of HBr are generated, what volume of 0.100 M H2SO4 is needed to produce this much HBr? lewis structure for SbF5? The Correct Answer and Explanation is : Let’s start with the first problem: Given balanced equation: [\\text{CaBr}_2(aq) + \\text{H}_2\\text{SO}_4(aq) […]<\/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-195269","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/195269","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=195269"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/195269\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=195269"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=195269"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=195269"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}