{"id":40614,"date":"2025-06-27T18:51:22","date_gmt":"2025-06-27T18:51:22","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=40614"},"modified":"2025-06-27T18:51:24","modified_gmt":"2025-06-27T18:51:24","slug":"limiting-reactant-and-percent-yield-worksheet-consider-the-following-reaction","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/limiting-reactant-and-percent-yield-worksheet-consider-the-following-reaction\/","title":{"rendered":"Limiting Reactant and Percent Yield Worksheet Consider the following reaction"},"content":{"rendered":"\n
Limiting Reactant and Percent Yield Worksheet Consider the following reaction: HCl + H2O -> H3O+ + Cl- If 21 moles of HCl react with 6.96 moles of H2O, what are the limiting reactants? Consider the following reaction: FeS2 + 10O2 -> Fe2O3 + 8SO2 When 26.62 moles of FeS2 reacts with 59.44 moles of O2, what are the limiting and excess reactants? Consider the following reaction: Si + 2N2 -> Si3N4 When 600 g of Si reacts with 500 g of N2, what are the limiting and excess reactants? Given the following equation: 2Al(SO3)3 + 6NaOH -> 3Na2SO3 + 2Al(OH)3 If 10.0 g of Al(SO3)3 reacted with 10.0 g of NaOH, determine the limiting and excess reactants<\/p>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n
Limiting Reactant and Percent Yield Worksheet Solutions<\/strong><\/h3>\n\n\n\n\n\n\n\n
Problem 1:<\/strong> Reaction:<\/strong> HCl + H\u2082O \u2192 H\u2083O\u207a + Cl\u207b Given:<\/strong> 21 moles of HCl and 6.96 moles of H\u2082O<\/p>\n\n\n\n
This reaction has a 1:1 molar ratio between HCl and H\u2082O. That means 1 mole of HCl reacts with 1 mole of H\u2082O.<\/p>\n\n\n\n
\n
Required amount of H\u2082O for 21 moles of HCl:<\/strong> 21 moles HCl \u00d7 (1 mole H\u2082O \/ 1 mole HCl) = 21 moles H\u2082O<\/li>\n<\/ul>\n\n\n\n
Available:<\/strong> 6.96 moles H\u2082O Conclusion:<\/strong> H\u2082O is the limiting reactant because there is not enough to react with all the HCl.<\/p>\n\n\n\n\n\n\n\n
Problem 2:<\/strong> Reaction:<\/strong> 4FeS\u2082 + 11O\u2082 \u2192 2Fe\u2082O\u2083 + 8SO\u2082<\/p>\n\n\n\n
Given:<\/strong> 26.62 moles FeS\u2082 and 59.44 moles O\u2082<\/p>\n\n\n\n
Conclusion:<\/strong> Al(SO\u2083)\u2083 is the limiting reactant. NaOH is in excess.<\/p>\n\n\n\n\n\n\n\n
Explanation Summary<\/strong><\/h3>\n\n\n\n
The limiting reactant is the substance that gets completely consumed first during a chemical reaction, limiting the amount of product formed. To identify the limiting reactant, you compare the actual mole ratio of reactants with the required stoichiometric ratio from the balanced equation. The reactant that cannot meet the required amount is the limiting reactant. Once the limiting reactant is used up, the reaction stops, even if other reactants are still present. In each problem above, careful mole calculations and conversions help determine which substance is limiting and which is in excess, based on the balanced chemical equation and given quantities.<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"
Limiting Reactant and Percent Yield Worksheet Consider the following reaction: HCl + H2O -> H3O+ + Cl- If 21 moles of HCl react with 6.96 moles of H2O, what are the limiting reactants? Consider the following reaction: FeS2 + 10O2 -> Fe2O3 + 8SO2 When 26.62 moles of FeS2 reacts with 59.44 moles of O2, […]<\/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-40614","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/40614","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=40614"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/40614\/revisions"}],"predecessor-version":[{"id":40624,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/40614\/revisions\/40624"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=40614"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=40614"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=40614"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
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