{"id":45702,"date":"2025-07-01T10:29:04","date_gmt":"2025-07-01T10:29:04","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=45702"},"modified":"2025-07-01T10:29:06","modified_gmt":"2025-07-01T10:29:06","slug":"caoh-is-used-to-remove-co-during-general-anesthesia-to-prevent-co-poisoning","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/caoh-is-used-to-remove-co-during-general-anesthesia-to-prevent-co-poisoning\/","title":{"rendered":"Ca(OH)? is used to remove CO? during general anesthesia to prevent CO? poisoning"},"content":{"rendered":"\n

Ca(OH)? is used to remove CO? during general anesthesia to prevent CO? poisoning: Ca(OH)?(s) + CO?(g) ? CaCO?(s) + H?O(l) a. If 12.0 g of CO? are exhaled and absorbed by 12.0 g Ca(OH)?, how many grams of CO? and Ca(OH)? are left over at the end of the reaction? b. To ensure that all the CO? is captured and removed, an excess amount of Ca(OH)? is generally used. If we want 12.0 g of CO? to only consume 10% of the available Ca(OH)?, what mass of Ca(OH)? should be used?<\/p>\n\n\n\n

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

Let’s break this down step by step.<\/p>\n\n\n\n

The reaction is: Ca(OH)2(s)+CO2(g)\u2192CaCO3(s)+H2O(l)\\text{Ca(OH)}_2 (s) + \\text{CO}_2 (g) \\rightarrow \\text{CaCO}_3 (s) + \\text{H}_2\\text{O} (l)Ca(OH)2\u200b(s)+CO2\u200b(g)\u2192CaCO3\u200b(s)+H2\u200bO(l)<\/p>\n\n\n\n

Part (a): Grams of CO\u2082 and Ca(OH)\u2082 left over<\/h3>\n\n\n\n

To determine the amount of CO\u2082 and Ca(OH)\u2082 left over, we first need to balance the equation and calculate how much of each substance will react.<\/p>\n\n\n\n

Step 1: Molar masses of the compounds involved<\/h4>\n\n\n\n
    \n
  • Molar mass of Ca(OH)\u2082<\/strong> = 74.09 g\/mol<\/li>\n\n\n\n
  • Molar mass of CO\u2082<\/strong> = 44.01 g\/mol<\/li>\n<\/ul>\n\n\n\n

    Step 2: Convert the given masses to moles<\/h4>\n\n\n\n

    We are given 12.0 g of each substance, so we can calculate the moles.<\/p>\n\n\n\n

    For Ca(OH)\u2082<\/strong>: moles of Ca(OH)2=12.0\u2009g74.09\u2009g\/mol=0.162\u2009mol\\text{moles of Ca(OH)}_2 = \\frac{12.0 \\, \\text{g}}{74.09 \\, \\text{g\/mol}} = 0.162 \\, \\text{mol}moles of Ca(OH)2\u200b=74.09g\/mol12.0g\u200b=0.162mol<\/p>\n\n\n\n

    For CO\u2082<\/strong>: moles of CO2=12.0\u2009g44.01\u2009g\/mol=0.273\u2009mol\\text{moles of CO}_2 = \\frac{12.0 \\, \\text{g}}{44.01 \\, \\text{g\/mol}} = 0.273 \\, \\text{mol}moles of CO2\u200b=44.01g\/mol12.0g\u200b=0.273mol<\/p>\n\n\n\n

    Step 3: Identify the limiting reagent<\/h4>\n\n\n\n

    From the balanced equation, we see that 1 mole of Ca(OH)\u2082 reacts with 1 mole of CO\u2082<\/strong>. So, the limiting reagent will be the substance that is present in the lesser amount of moles.<\/p>\n\n\n\n

      \n
    • We have 0.162 mol of Ca(OH)\u2082<\/strong> and 0.273 mol of CO\u2082<\/strong>.<\/li>\n\n\n\n
    • Since Ca(OH)\u2082<\/strong> has fewer moles, it is the limiting reagent.<\/li>\n<\/ul>\n\n\n\n

      Step 4: Calculate the amount of CO\u2082 that reacts<\/h4>\n\n\n\n

      Since Ca(OH)\u2082<\/strong> is the limiting reagent, all of it will react with CO\u2082. For every 1 mole of Ca(OH)\u2082, 1 mole of CO\u2082 will be consumed.<\/p>\n\n\n\n

      Thus, 0.162 mol of CO\u2082<\/strong> will react with 0.162 mol of Ca(OH)\u2082<\/strong>.<\/p>\n\n\n\n

      Step 5: Calculate the remaining CO\u2082<\/h4>\n\n\n\n

      We started with 0.273 mol of CO\u2082<\/strong>, and 0.162 mol<\/strong> will react, so the remaining CO\u2082<\/strong> will be: Remaining CO2=0.273\u2009mol\u22120.162\u2009mol=0.111\u2009mol\\text{Remaining CO}_2 = 0.273 \\, \\text{mol} – 0.162 \\, \\text{mol} = 0.111 \\, \\text{mol}Remaining CO2\u200b=0.273mol\u22120.162mol=0.111mol<\/p>\n\n\n\n

      Now, convert this back to grams: Remaining CO2=0.111\u2009mol\u00d744.01\u2009g\/mol=4.88\u2009g\\text{Remaining CO}_2 = 0.111 \\, \\text{mol} \\times 44.01 \\, \\text{g\/mol} = 4.88 \\, \\text{g}Remaining CO2\u200b=0.111mol\u00d744.01g\/mol=4.88g<\/p>\n\n\n\n

      Step 6: Calculate the remaining Ca(OH)\u2082<\/h4>\n\n\n\n

      Since 0.162 mol of Ca(OH)\u2082<\/strong> reacts with 0.162 mol of CO\u2082<\/strong>, and we started with 0.162 mol of Ca(OH)\u2082<\/strong>, all of it will be consumed.<\/p>\n\n\n\n

      Therefore, no Ca(OH)\u2082 will be left over.<\/p>\n\n\n\n

      Part (b): Mass of Ca(OH)\u2082 for 10% reaction<\/h3>\n\n\n\n

      We want to ensure that 12.0 g of CO\u2082<\/strong> consumes only 10% of the available Ca(OH)\u2082<\/strong>.<\/p>\n\n\n\n

      Step 1: Calculate moles of CO\u2082<\/h4>\n\n\n\n

      We know from earlier: moles of CO2=12.0\u2009g44.01\u2009g\/mol=0.273\u2009mol\\text{moles of CO}_2 = \\frac{12.0 \\, \\text{g}}{44.01 \\, \\text{g\/mol}} = 0.273 \\, \\text{mol}moles of CO2\u200b=44.01g\/mol12.0g\u200b=0.273mol<\/p>\n\n\n\n

      Step 2: Calculate the moles of Ca(OH)\u2082 required<\/h4>\n\n\n\n

      The balanced equation tells us that 1 mole of Ca(OH)\u2082 reacts with 1 mole of CO\u2082<\/strong>, so 0.273 mol of CO\u2082<\/strong> will need 0.273 mol of Ca(OH)\u2082<\/strong>.<\/p>\n\n\n\n

      Step 3: Find the total moles of Ca(OH)\u2082<\/h4>\n\n\n\n

      If 0.273 mol of Ca(OH)\u2082<\/strong> is only 10% of the total amount available, then the total moles of Ca(OH)\u2082 needed will be: Total moles of Ca(OH)2=0.273\u2009mol0.10=2.73\u2009mol\\text{Total moles of Ca(OH)}_2 = \\frac{0.273 \\, \\text{mol}}{0.10} = 2.73 \\, \\text{mol}Total moles of Ca(OH)2\u200b=0.100.273mol\u200b=2.73mol<\/p>\n\n\n\n

      Step 4: Convert moles of Ca(OH)\u2082 to mass<\/h4>\n\n\n\n

      Now, we calculate the mass of Ca(OH)\u2082<\/strong> required: Mass of Ca(OH)2=2.73\u2009mol\u00d774.09\u2009g\/mol=202.5\u2009g\\text{Mass of Ca(OH)}_2 = 2.73 \\, \\text{mol} \\times 74.09 \\, \\text{g\/mol} = 202.5 \\, \\text{g}Mass of Ca(OH)2\u200b=2.73mol\u00d774.09g\/mol=202.5g<\/p>\n\n\n\n

      Final Answers:<\/h3>\n\n\n\n
        \n
      • Part (a)<\/strong>: 4.88 g of CO\u2082 will be left over, and no Ca(OH)\u2082 will remain.<\/li>\n\n\n\n
      • Part (b)<\/strong>: To ensure that only 10% of Ca(OH)\u2082 is used to capture 12.0 g of CO\u2082, 202.5 g of Ca(OH)\u2082<\/strong> should be used.<\/li>\n<\/ul>\n\n\n\n
        \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

        Ca(OH)? is used to remove CO? during general anesthesia to prevent CO? poisoning: Ca(OH)?(s) + CO?(g) ? CaCO?(s) + H?O(l) a. If 12.0 g of CO? are exhaled and absorbed by 12.0 g Ca(OH)?, how many grams of CO? and Ca(OH)? are left over at the end of the reaction? b. To ensure that all […]<\/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-45702","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/45702","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=45702"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/45702\/revisions"}],"predecessor-version":[{"id":45708,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/45702\/revisions\/45708"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=45702"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=45702"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=45702"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}