{"id":195512,"date":"2025-02-28T10:25:54","date_gmt":"2025-02-28T10:25:54","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=195512"},"modified":"2025-02-28T10:25:57","modified_gmt":"2025-02-28T10:25:57","slug":"what-weight-of-manganese-is-present-in-2-58-g-of-mn3o4","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/28\/what-weight-of-manganese-is-present-in-2-58-g-of-mn3o4\/","title":{"rendered":"What weight of manganese is present in 2.58 g of Mn3O4"},"content":{"rendered":"\n

What weight of manganese is present in 2.58 g of Mn3O4?<\/p>\n\n\n\n

Calculate the gravimetric factor for: Substance sought (manganese) from Substance Weighed (Mn3O4).<\/p>\n\n\n\n

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

Step 1: Understanding Mn\u2083O\u2084 Composition<\/strong><\/h3>\n\n\n\n

Manganese(II,III) oxide ((Mn_3O_4)) consists of both (Mn^{2+}) and (Mn^{3+}) ions. The molecular weight of (Mn_3O_4) is calculated as:<\/p>\n\n\n\n

    \n
  • Atomic mass of Mn = 54.938 g\/mol<\/strong><\/li>\n\n\n\n
  • Atomic mass of O = 16.00 g\/mol<\/strong><\/li>\n<\/ul>\n\n\n\n

    [
    \\text{Molar mass of } Mn_3O_4 = (3 \\times 54.938) + (4 \\times 16.00) = 164.81 \\text{ g\/mol}
    ]<\/p>\n\n\n\n

    Step 2: Gravimetric Factor Calculation<\/strong><\/h3>\n\n\n\n

    The gravimetric factor is the ratio of the molar mass of the substance sought (manganese) to the substance weighed ((Mn_3O_4)).<\/p>\n\n\n\n

    [
    \\text{Gravimetric Factor} = \\frac{\\text{Mass of Mn in Mn}_3O_4}{\\text{Molar Mass of Mn}_3O_4}
    ]<\/p>\n\n\n\n

    The total mass of manganese in (Mn_3O_4):<\/p>\n\n\n\n

    [
    3 \\times 54.938 = 164.814 \\text{ g\/mol}
    ]<\/p>\n\n\n\n

    Thus, the gravimetric factor is:<\/p>\n\n\n\n

    [
    \\frac{164.814}{164.81} = 0.9999 \\approx 1.00
    ]<\/p>\n\n\n\n

    Step 3: Determining the Mass of Manganese in 2.58 g of Mn\u2083O\u2084<\/strong><\/h3>\n\n\n\n

    Now, using the gravimetric factor:<\/p>\n\n\n\n

    [
    \\text{Mass of Mn} = 2.58 \\times 0.9999 = 2.58 \\text{ g}
    ]<\/p>\n\n\n\n

    Conclusion<\/strong><\/h3>\n\n\n\n
      \n
    • Weight of manganese in 2.58 g of Mn\u2083O\u2084 is 2.58 g<\/strong><\/li>\n\n\n\n
    • Gravimetric factor for Mn from Mn\u2083O\u2084 is approximately 1.00<\/strong><\/li>\n<\/ul>\n\n\n\n

      Now, I’ll generate an image summarizing this calculation.<\/p>\n\n\n\n

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

      Here is the infographic summarizing the calculation of manganese weight in Mn\u2083O\u2084. Let me know if you need any modifications or further explanations!<\/p>\n","protected":false},"excerpt":{"rendered":"

      What weight of manganese is present in 2.58 g of Mn3O4? Calculate the gravimetric factor for: Substance sought (manganese) from Substance Weighed (Mn3O4). The Correct Answer and Explanation is : Step 1: Understanding Mn\u2083O\u2084 Composition Manganese(II,III) oxide ((Mn_3O_4)) consists of both (Mn^{2+}) and (Mn^{3+}) ions. The molecular weight of (Mn_3O_4) is calculated as: [\\text{Molar mass […]<\/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-195512","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/195512","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=195512"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/195512\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=195512"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=195512"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=195512"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}