{"id":250775,"date":"2025-07-10T12:14:46","date_gmt":"2025-07-10T12:14:46","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=250775"},"modified":"2025-07-10T12:14:50","modified_gmt":"2025-07-10T12:14:50","slug":"how-many-grams-of-water-are-produced-if-20-0g-of-hydrogen-peroxide-decompose","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/10\/how-many-grams-of-water-are-produced-if-20-0g-of-hydrogen-peroxide-decompose\/","title":{"rendered":"How many grams of water are produced if 20.0g of Hydrogen Peroxide decompose"},"content":{"rendered":"\n

Given the reaction 2H2O2 \u00e2\u2020\u2019 O2 + 2H2O, how many grams of water are produced if 20.0g of Hydrogen Peroxide decompose? 2. Corrected_text: Given the reaction 3Fe + 4H2O \u00e2\u2020\u2019 Fe3O4 + 4H2, how many grams of water are needed to react with 42.0g of Iron? 3. Corrected_text: Given the reaction 2Al + 6HCl \u00e2\u2020\u2019 2AlCl3 + 3H2, how many grams of Aluminum Chloride are produced if 4.80 moles of HCl are consumed<\/p>\n\n\n\n

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

Let’s break down each problem step by step:<\/p>\n\n\n\n

1. Hydrogen Peroxide Decomposition Reaction<\/strong>:<\/h3>\n\n\n\n

Given reaction<\/strong>:
2H2O2\u2192O2+2H2O2H_2O_2 \\rightarrow O_2 + 2H_2O2H2\u200bO2\u200b\u2192O2\u200b+2H2\u200bO<\/p>\n\n\n\n

Given<\/strong>: 20.0 g of hydrogen peroxide (H\u2082O\u2082)<\/p>\n\n\n\n

We need to calculate how many grams of water (H\u2082O) are produced.<\/p>\n\n\n\n

Step 1: Calculate the moles of H\u2082O\u2082<\/h4>\n\n\n\n

The molar mass of H\u2082O\u2082 is calculated as:
2\u00d7(1.008\u2009g\/mol\u2009(for H))+2\u00d7(15.999\u2009g\/mol\u2009(for O))=34.014\u2009g\/mol2 \\times (1.008 \\, \\text{g\/mol} \\, \\text{(for H)}) + 2 \\times (15.999 \\, \\text{g\/mol} \\, \\text{(for O)}) = 34.014 \\, \\text{g\/mol}2\u00d7(1.008g\/mol(for H))+2\u00d7(15.999g\/mol(for O))=34.014g\/mol<\/p>\n\n\n\n

Now, calculate the moles of H\u2082O\u2082 in 20.0 g: moles of H2O2=20.0\u2009g34.014\u2009g\/mol\u22480.588\u2009mol\\text{moles of H}_2\\text{O}_2 = \\frac{20.0 \\, \\text{g}}{34.014 \\, \\text{g\/mol}} \\approx 0.588 \\, \\text{mol}moles of H2\u200bO2\u200b=34.014g\/mol20.0g\u200b\u22480.588mol<\/p>\n\n\n\n

Step 2: Use stoichiometry to find moles of H\u2082O produced<\/h4>\n\n\n\n

From the balanced chemical equation, we see that 2 moles of H\u2082O\u2082 produce 2 moles of H\u2082O. Thus, the ratio is 1:1. Therefore, the moles of water produced are the same as the moles of H\u2082O\u2082.<\/p>\n\n\n\n

So, 0.588 mol of H\u2082O\u2082 will produce 0.588 mol of H\u2082O.<\/p>\n\n\n\n

Step 3: Convert moles of H\u2082O to grams<\/h4>\n\n\n\n

The molar mass of water (H\u2082O) is:
2\u00d7(1.008\u2009g\/mol)+1\u00d7(15.999\u2009g\/mol)=18.016\u2009g\/mol2 \\times (1.008 \\, \\text{g\/mol}) + 1 \\times (15.999 \\, \\text{g\/mol}) = 18.016 \\, \\text{g\/mol}2\u00d7(1.008g\/mol)+1\u00d7(15.999g\/mol)=18.016g\/mol<\/p>\n\n\n\n

Now, calculate the mass of water produced: mass of H2O=0.588\u2009mol\u00d718.016\u2009g\/mol=10.6\u2009g\\text{mass of H}_2\\text{O} = 0.588 \\, \\text{mol} \\times 18.016 \\, \\text{g\/mol} = 10.6 \\, \\text{g}mass of H2\u200bO=0.588mol\u00d718.016g\/mol=10.6g<\/p>\n\n\n\n

So, 10.6 grams<\/strong> of water are produced.<\/p>\n\n\n\n

\n\n\n\n

2. Iron and Water Reaction<\/strong>:<\/h3>\n\n\n\n

Given reaction<\/strong>:
3Fe+4H2O\u2192Fe3O4+4H23Fe + 4H_2O \\rightarrow Fe_3O_4 + 4H_23Fe+4H2\u200bO\u2192Fe3\u200bO4\u200b+4H2\u200b<\/p>\n\n\n\n

Given<\/strong>: 42.0 g of Iron (Fe)<\/p>\n\n\n\n

We need to calculate how many grams of water (H\u2082O) are needed to react with 42.0 g of iron.<\/p>\n\n\n\n

Step 1: Calculate the moles of iron (Fe)<\/h4>\n\n\n\n

The molar mass of iron (Fe) is:
55.845\u2009g\/mol55.845 \\, \\text{g\/mol}55.845g\/mol<\/p>\n\n\n\n

Now, calculate the moles of Fe in 42.0 g: moles of Fe=42.0\u2009g55.845\u2009g\/mol\u22480.752\u2009mol\\text{moles of Fe} = \\frac{42.0 \\, \\text{g}}{55.845 \\, \\text{g\/mol}} \\approx 0.752 \\, \\text{mol}moles of Fe=55.845g\/mol42.0g\u200b\u22480.752mol<\/p>\n\n\n\n

Step 2: Use stoichiometry to find moles of H\u2082O needed<\/h4>\n\n\n\n

From the balanced chemical equation, we see that 3 moles of Fe react with 4 moles of H\u2082O. Thus, the ratio of Fe to H\u2082O is 3:4.<\/p>\n\n\n\n

Now, calculate the moles of water needed: moles of H2O=43\u00d70.752\u2009mol Fe\u22481.003\u2009mol H2O\\text{moles of H}_2\\text{O} = \\frac{4}{3} \\times 0.752 \\, \\text{mol Fe} \\approx 1.003 \\, \\text{mol H}_2\\text{O}moles of H2\u200bO=34\u200b\u00d70.752mol Fe\u22481.003mol H2\u200bO<\/p>\n\n\n\n

Step 3: Convert moles of H\u2082O to grams<\/h4>\n\n\n\n

The molar mass of water (H\u2082O) is:
18.016\u2009g\/mol18.016 \\, \\text{g\/mol}18.016g\/mol<\/p>\n\n\n\n

Now, calculate the mass of water needed: mass of H2O=1.003\u2009mol\u00d718.016\u2009g\/mol\u224818.1\u2009g\\text{mass of H}_2\\text{O} = 1.003 \\, \\text{mol} \\times 18.016 \\, \\text{g\/mol} \\approx 18.1 \\, \\text{g}mass of H2\u200bO=1.003mol\u00d718.016g\/mol\u224818.1g<\/p>\n\n\n\n

So, 18.1 grams<\/strong> of water are needed to react with 42.0 g of iron.<\/p>\n\n\n\n

\n\n\n\n

3. Aluminum Chloride Production<\/strong>:<\/h3>\n\n\n\n

Given reaction<\/strong>:
2Al+6HCl\u21922AlCl3+3H22Al + 6HCl \\rightarrow 2AlCl_3 + 3H_22Al+6HCl\u21922AlCl3\u200b+3H2\u200b<\/p>\n\n\n\n

Given<\/strong>: 4.80 moles of HCl<\/p>\n\n\n\n

We need to calculate how many grams of Aluminum Chloride (AlCl\u2083) are produced.<\/p>\n\n\n\n

Step 1: Use stoichiometry to find moles of AlCl\u2083 produced<\/h4>\n\n\n\n

From the balanced chemical equation, we see that 6 moles of HCl produce 2 moles of AlCl\u2083. Thus, the ratio of HCl to AlCl\u2083 is 6:2 or 3:1.<\/p>\n\n\n\n

Now, calculate the moles of AlCl\u2083 produced: moles of AlCl3=26\u00d74.80\u2009mol HCl=1.60\u2009mol AlCl3\\text{moles of AlCl}_3 = \\frac{2}{6} \\times 4.80 \\, \\text{mol HCl} = 1.60 \\, \\text{mol AlCl}_3moles of AlCl3\u200b=62\u200b\u00d74.80mol HCl=1.60mol AlCl3\u200b<\/p>\n\n\n\n

Step 2: Convert moles of AlCl\u2083 to grams<\/h4>\n\n\n\n

The molar mass of AlCl\u2083 is:
26.98\u2009g\/mol\u2009(for Al)+3\u00d735.45\u2009g\/mol\u2009(for Cl)=133.34\u2009g\/mol26.98 \\, \\text{g\/mol} \\, \\text{(for Al)} + 3 \\times 35.45 \\, \\text{g\/mol} \\, \\text{(for Cl)} = 133.34 \\, \\text{g\/mol}26.98g\/mol(for Al)+3\u00d735.45g\/mol(for Cl)=133.34g\/mol<\/p>\n\n\n\n

Now, calculate the mass of AlCl\u2083 produced: mass of AlCl3=1.60\u2009mol\u00d7133.34\u2009g\/mol=213.34\u2009g\\text{mass of AlCl}_3 = 1.60 \\, \\text{mol} \\times 133.34 \\, \\text{g\/mol} = 213.34 \\, \\text{g}mass of AlCl3\u200b=1.60mol\u00d7133.34g\/mol=213.34g<\/p>\n\n\n\n

So, 213.34 grams<\/strong> of Aluminum Chloride are produced when 4.80 moles of HCl are consumed.<\/p>\n\n\n\n

\n\n\n\n

Summary of Answers:<\/h3>\n\n\n\n
    \n
  1. Water produced<\/strong>: 10.6 grams<\/li>\n\n\n\n
  2. Water needed<\/strong>: 18.1 grams<\/li>\n\n\n\n
  3. Aluminum Chloride produced<\/strong>: 213.34 grams<\/li>\n<\/ol>\n\n\n\n
    \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

    Given the reaction 2H2O2 \u00e2\u2020\u2019 O2 + 2H2O, how many grams of water are produced if 20.0g of Hydrogen Peroxide decompose? 2. Corrected_text: Given the reaction 3Fe + 4H2O \u00e2\u2020\u2019 Fe3O4 + 4H2, how many grams of water are needed to react with 42.0g of Iron? 3. Corrected_text: Given the reaction 2Al + 6HCl \u00e2\u2020\u2019 […]<\/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-250775","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/250775","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=250775"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/250775\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=250775"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=250775"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=250775"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}