If copper (I) sulfide is partially roasted in air (a reaction with O2), copper (I) sulfite [Cu2SO3] is first formed. Subsequently, upon heating, the copper (I) sulfite thermally decomposes to copper (I) oxide and sulfur dioxide. Write balanced chemical equations for these two reactions.<\/p>\n\n\n\n
A<\/p>\n\n\n\n
B<\/p>\n\n\n\n
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n Alright, let\u2019s tackle it carefully.<\/p>\n\n\n\n First Reaction (A): Formation of Copper(I) Sulfite (Cu\u2082SO\u2083)<\/strong> Balanced Equation: Second Reaction (B): Decomposition of Copper(I) Sulfite<\/strong> Balanced Equation: In metallurgy and chemical processes, roasting refers to heating ores strongly in the presence of air. Here, copper(I) sulfide (Cu\u2082S), an important copper ore (also called chalcocite), is partially roasted<\/strong> in air. Partial roasting means that there isn’t an excess of oxygen; thus, complete conversion to oxides doesn’t happen immediately.<\/p>\n\n\n\n Initially, Cu\u2082S reacts with O\u2082 to form copper(I) sulfite (Cu\u2082SO\u2083)<\/strong> rather than fully oxidizing to copper(I) oxide. This is because, under limited oxygen, sulfur doesn’t fully oxidize to sulfur dioxide (SO\u2082) but forms an intermediate sulfite (SO\u2083\u00b2\u207b). The reaction conserves the copper\u2019s +1 oxidation state throughout, forming Cu\u2082SO\u2083.<\/p>\n\n\n\n However, copper(I) sulfite is thermally unstable<\/strong>. Upon further heating, it decomposes<\/strong> into copper(I) oxide (Cu\u2082O)<\/strong> and releases sulfur dioxide (SO\u2082)<\/strong> gas. Here, the sulfite ion (SO\u2083\u00b2\u207b) loses one oxygen atom, forming SO\u2082. The copper part, staying in the +1 oxidation state, ends up as Cu\u2082O, a red solid.<\/p>\n\n\n\n Both reactions are important in copper extraction processes. The first step helps prepare the ore for easier removal of sulfur, and the second ensures that sulfur is expelled as a gas (SO\u2082), leaving behind metallic copper compounds that can be reduced to pure copper. Also, controlling oxygen supply is crucial; too much oxygen would over-oxidize copper to CuO (copper(II) oxide).<\/p>\n\n\n\n Thus, these two steps show how careful temperature and air control help transform sulfide ores into usable copper products.<\/p>\n\n\n\n If copper (I) sulfide is partially roasted in air (a reaction with O2), copper (I) sulfite [Cu2SO3] is first formed. Subsequently, upon heating, the copper (I) sulfite thermally decomposes to copper (I) oxide and sulfur dioxide. Write balanced chemical equations for these two reactions. A B The correct answer and explanation is : Alright, let\u2019s […]<\/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-208684","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/208684","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=208684"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/208684\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=208684"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=208684"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=208684"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
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When copper(I) sulfide (Cu\u2082S) is partially roasted<\/strong> (heated in limited air), it reacts with oxygen (O\u2082) to form copper(I) sulfite (Cu\u2082SO\u2083).<\/p>\n\n\n\n
[
\\text{A:} \\quad \\text{Cu}_2\\text{S} + \\text{O}_2 \\rightarrow \\text{Cu}_2\\text{SO}_3
]<\/p>\n\n\n\n
\n\n\n\n
When the copper(I) sulfite (Cu\u2082SO\u2083) is heated further, it thermally decomposes<\/strong> into copper(I) oxide (Cu\u2082O) and sulfur dioxide (SO\u2082).<\/p>\n\n\n\n
[
\\text{B:} \\quad \\text{Cu}_2\\text{SO}_3 \\rightarrow \\text{Cu}_2\\text{O} + \\text{SO}_2
]<\/p>\n\n\n\n
\n\n\n\nExplanation (Around 300 Words)<\/h3>\n\n\n\n
\n","protected":false},"excerpt":{"rendered":"