{"id":29440,"date":"2025-06-21T04:34:38","date_gmt":"2025-06-21T04:34:38","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=29440"},"modified":"2025-06-21T04:34:41","modified_gmt":"2025-06-21T04:34:41","slug":"sodium-hydroxide-reacts-with-oxalic-acid-as-shown-in-the-reaction-below-h2c2o4s-2-naohaq-a%e2%80%a0-na2c2o4aq-2-h2ol","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/sodium-hydroxide-reacts-with-oxalic-acid-as-shown-in-the-reaction-below-h2c2o4s-2-naohaq-a%e2%80%a0-na2c2o4aq-2-h2ol\/","title":{"rendered":"Sodium hydroxide reacts with oxalic acid as shown in the reaction below: H2C2O4(s) + 2 NaOH(aq) \u00e2\u2020\u2019 Na2C2O4(aq) + 2 H2O(l)"},"content":{"rendered":"\n
Sodium hydroxide reacts with oxalic acid as shown in the reaction below: H2C2O4(s) + 2 NaOH(aq) \u00e2\u2020\u2019 Na2C2O4(aq) + 2 H2O(l) a) If 0.312 moles of sodium hydroxide are reacted with 4.59 g of oxalic acid, what mass of the excess reactant remains after the reaction is complete? Be clear about which species you are calculating. b) The percent yield of this reaction is found to be 88.6%. If you require 4.65 moles of sodium oxalate for a subsequent reaction, how many kilograms of oxalic acid must be added to excess sodium hydroxide?<\/p>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n
a) Limiting Reactant and Excess Mass Remaining<\/strong><\/h3>\n\n\n\n
From the stoichiometry:<\/strong> 1 mol oxalic acid reacts with 2 mol NaOH So, 0.0510 mol oxalic acid requires:0.0510\u00d72=0.102 mol NaOH0.0510 \\times 2 = 0.102\\ \\text{mol NaOH}0.0510\u00d72=0.102 mol NaOH<\/p>\n\n\n\n
We have 0.312 mol NaOH, which is more than required. Thus, oxalic acid is the limiting reactant<\/strong>, and NaOH is in excess.<\/strong><\/p>\n\n\n\n
Mass of excess NaOH:<\/strong> Molar mass of NaOH = 22.99 + 15.999 + 1.008 = 39.997 g\/mol0.210\u00d739.997=8.40 g0.210 \\times 39.997 = 8.40\\ \\text{g}0.210\u00d739.997=8.40 g<\/p>\n\n\n\n
Answer (a):<\/strong> Excess species:<\/strong> Sodium hydroxide Mass remaining after reaction:<\/strong> 8.40 g<\/p>\n\n\n\n\n\n\n\n
b) Required Oxalic Acid for 4.65 mol Na\u2082C\u2082O\u2084 with 88.6% yield<\/strong><\/h3>\n\n\n\n
From the balanced equation: 1 mol Na\u2082C\u2082O\u2084 is produced from 1 mol oxalic acid<\/p>\n\n\n\n
So ideally, to get 4.65 mol of Na\u2082C\u2082O\u2084:<\/strong>4.65 mol H2C2O4required(theoretical)4.65\\ \\text{mol H}_2\\text{C}_2\\text{O}_4 required (theoretical)4.65 mol H2\u200bC2\u200bO4\u200brequired(theoretical)<\/p>\n\n\n\n
Since the reaction is only 88.6% efficient, use:Actual mol needed=4.650.886\u22485.25 mol\\text{Actual mol needed} = \\frac{4.65}{0.886} \\approx 5.25\\ \\text{mol}Actual mol needed=0.8864.65\u200b\u22485.25 mol<\/p>\n\n\n\n
Mass of oxalic acid needed:<\/strong>5.25 mol\u00d790.03 g\/mol=472.66 g5.25\\ \\text{mol} \\times 90.03\\ \\text{g\/mol} = 472.66\\ \\text{g}5.25 mol\u00d790.03 g\/mol=472.66 g<\/p>\n\n\n\n
Answer (b):<\/strong> You must add 0.4727 kilograms of oxalic acid to excess NaOH<\/strong> to obtain 4.65 mol of sodium oxalate at 88.6% yield.<\/p>\n\n\n\n\n\n\n\n
This problem demonstrates how stoichiometric coefficients and percent yield guide reagent requirements and identify leftover materials. The limiting reactant determines the reaction extent, while excess reactants help ensure completion. Percent yield corrects for real-life inefficiencies.<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"
Sodium hydroxide reacts with oxalic acid as shown in the reaction below: H2C2O4(s) + 2 NaOH(aq) \u00e2\u2020\u2019 Na2C2O4(aq) + 2 H2O(l) a) If 0.312 moles of sodium hydroxide are reacted with 4.59 g of oxalic acid, what mass of the excess reactant remains after the reaction is complete? Be clear about which species you are […]<\/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-29440","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/29440","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=29440"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/29440\/revisions"}],"predecessor-version":[{"id":29442,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/29440\/revisions\/29442"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=29440"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=29440"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=29440"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
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