{"id":22556,"date":"2025-06-16T19:55:49","date_gmt":"2025-06-16T19:55:49","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=22556"},"modified":"2025-06-16T19:55:57","modified_gmt":"2025-06-16T19:55:57","slug":"a-hydrate-of-cocl2-with-a-mass-of-6-00-g-is-heated-strongly","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/a-hydrate-of-cocl2-with-a-mass-of-6-00-g-is-heated-strongly\/","title":{"rendered":"A hydrate of CoCl2 with a mass of 6.00 g is heated strongly"},"content":{"rendered":"\n<p>A hydrate of CoCl2 with a mass of 6.00 g is heated strongly. After cooling, the mass of the anhydrate is 3.27 g. <\/p>\n\n\n\n<p>a. How many grams of H2O were lost from the hydrate? <\/p>\n\n\n\n<p>b. What is the % water in the hydrate? <\/p>\n\n\n\n<p>c. What is the formula of the CoCl2 hydrate? <\/p>\n\n\n\n<p>d. Write the equation for the dehydration of CoCl2 hydrate.<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\"><strong>The correct answer and explanation is:<\/strong><\/mark><\/p>\n\n\n\n<p>Sure! Let&#8217;s analyze the problem step-by-step.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Given:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mass of hydrate (CoCl\u2082\u00b7xH\u2082O) = 6.00 g<\/li>\n\n\n\n<li>Mass of anhydrate (CoCl\u2082) after heating = 3.27 g<\/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\">a. How many grams of H\u2082O were lost from the hydrate?<\/h3>\n\n\n\n<p>The water lost = Mass of hydrate \u2212 Mass of anhydrate<br>= 6.00 g \u2212 3.27 g = <strong>2.73 g<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">b. What is the % water in the hydrate?<\/h3>\n\n\n\n<p>% water = (mass of water \/ mass of hydrate) \u00d7 100<br>= (2.73 g \/ 6.00 g) \u00d7 100 = <strong>45.5%<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">c. What is the formula of the CoCl\u2082 hydrate?<\/h3>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>We need to find the number of water molecules, <em>x<\/em>, in CoCl\u2082\u00b7xH\u2082O.<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Calculate moles of anhydrate (CoCl\u2082):<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Molar mass of CoCl\u2082 =<br>Co = 58.93 g\/mol<br>Cl = 35.45 g\/mol \u00d7 2 = 70.90 g\/mol<br>Total = 58.93 + 70.90 = 129.83 g\/mol<\/li>\n<\/ul>\n\n\n\n<p>Moles CoCl\u2082 = mass \/ molar mass = 3.27 g \/ 129.83 g\/mol \u2248 0.02517 mol<\/p>\n\n\n\n<ol start=\"2\" class=\"wp-block-list\">\n<li><strong>Calculate moles of water lost:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Molar mass of H\u2082O = 18.02 g\/mol<br>Moles H\u2082O = 2.73 g \/ 18.02 g\/mol \u2248 0.1515 mol<\/li>\n<\/ul>\n\n\n\n<ol start=\"3\" class=\"wp-block-list\">\n<li><strong>Find mole ratio of H\u2082O to CoCl\u2082:<\/strong><\/li>\n<\/ol>\n\n\n\n<p>Ratio = moles H\u2082O \/ moles CoCl\u2082<br>= 0.1515 \/ 0.02517 \u2248 6.02 \u2248 6<\/p>\n\n\n\n<p>Thus, the formula of the hydrate is <strong>CoCl\u2082\u00b76H\u2082O<\/strong>.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">d. Write the equation for the dehydration of CoCl\u2082 hydrate.<\/h3>\n\n\n\n<p>When the hydrate is heated, it loses water and forms anhydrous cobalt(II) chloride: CoCl2\u22c56H2O\u2192heatCoCl2+6H2O\\mathrm{CoCl_2 \\cdot 6H_2O \\xrightarrow{\\text{heat}} CoCl_2 + 6H_2O}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation (300 words):<\/h3>\n\n\n\n<p>Hydrates are compounds that contain water molecules integrated into their crystalline structure. Heating a hydrate usually causes the water molecules to evaporate, leaving behind the anhydrous salt.<\/p>\n\n\n\n<p>In this problem, a 6.00 g sample of cobalt(II) chloride hydrate (CoCl\u2082\u00b7xH\u2082O) is heated until all the water is lost, leaving 3.27 g of anhydrous CoCl\u2082. The difference in mass (6.00 g \u2212 3.27 g = 2.73 g) represents the mass of water lost.<\/p>\n\n\n\n<p>To determine the water percentage in the hydrate, the mass of water lost is divided by the original hydrate mass and multiplied by 100, giving 45.5%. This means nearly half of the mass of the hydrate is due to water.<\/p>\n\n\n\n<p>Next, to find the exact number of water molecules per formula unit (the value of x), we convert the masses to moles. The moles of anhydrous CoCl\u2082 are calculated using its molar mass (129.83 g\/mol), and the moles of water lost are calculated using water&#8217;s molar mass (18.02 g\/mol). The ratio of moles of water to moles of CoCl\u2082 comes out to approximately 6, indicating that each formula unit of cobalt(II) chloride hydrate contains 6 water molecules.<\/p>\n\n\n\n<p>Finally, the dehydration reaction can be written as: CoCl2\u22c56H2O\u2192heatCoCl2+6H2O\\mathrm{CoCl_2 \\cdot 6H_2O \\xrightarrow{\\text{heat}} CoCl_2 + 6H_2O}<\/p>\n\n\n\n<p>This reaction demonstrates the loss of water molecules upon heating, transforming the hydrate into its anhydrous form.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>A hydrate of CoCl2 with a mass of 6.00 g is heated strongly. After cooling, the mass of the anhydrate is 3.27 g. a. How many grams of H2O were lost from the hydrate? b. What is the % water in the hydrate? c. What is the formula of the CoCl2 hydrate? d. Write the [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-22556","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/22556","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=22556"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/22556\/revisions"}],"predecessor-version":[{"id":22557,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/22556\/revisions\/22557"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=22556"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=22556"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=22556"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}