{"id":186541,"date":"2025-01-25T09:00:15","date_gmt":"2025-01-25T09:00:15","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=186541"},"modified":"2025-01-25T09:00:18","modified_gmt":"2025-01-25T09:00:18","slug":"chm-151-hw-7","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/25\/chm-151-hw-7\/","title":{"rendered":"CHM 151: HW 7"},"content":{"rendered":"\n

CHM 151: HW 7 1. NH O2 NO H\u2082O 2. Al+ Cr2O3 Al2O3+ Cr 3. Ca3(PO4)2 + H2SO4 CaSO+ H3PO4 4. C7H14+ O2 CO2 + H2O 5. P4O10+ H\u2082O H3PO4 6. NaCl + H2SO4 HCI Na2SO4 7. Al+ H2SO4 Al2(SO4)3 H 8. Consider the reaction below. How many moles of FeS: will react with 12.2 moles of oxygen? 4 FeS2 + 11 02 2 Fe2O3 + 8 SO2 9. In the reaction below, 10.28 moles of Carbon dioxide are formed from 7.7 moles of the hydrocarbon. What is the percentage yield? 2 C4H10 13 028 CO2 10 H\u2082O<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

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

To help with your HW, I’ll break down each question and provide the relevant steps to find the answers.<\/p>\n\n\n\n

1. NH\u2082O + O\u2082 \u2192 NO + H\u2082O<\/strong><\/h3>\n\n\n\n

This is a redox reaction where nitrogen in NH\u2082O is oxidized to NO, and oxygen is reduced. To balance this, you’d need to ensure both mass and charge balance. The molecular weights of NH\u2082O, O\u2082, NO, and H\u2082O will help ensure that the atoms are conserved.<\/p>\n\n\n\n

2. Al + Cr\u2082O\u2083 \u2192 Al\u2082O\u2083 + Cr<\/strong><\/h3>\n\n\n\n

This is a redox reaction between aluminum and chromium oxide. Aluminum displaces chromium because it is more reactive. The stoichiometric coefficients should balance to ensure the conservation of atoms. The reaction would look like this after balancing:
[
6Al + Cr\u2082O\u2083 \\rightarrow 2Al\u2082O\u2083 + 2Cr
]<\/p>\n\n\n\n

3. Ca\u2083(PO\u2084)\u2082 + H\u2082SO\u2084 \u2192 CaSO\u2084 + H\u2083PO\u2084<\/strong><\/h3>\n\n\n\n

This is an acid-base reaction where calcium phosphate reacts with sulfuric acid. The products are calcium sulfate and phosphoric acid. The balanced equation will be:
[
Ca\u2083(PO\u2084)\u2082 + 3H\u2082SO\u2084 \u2192 3CaSO\u2084 + 2H\u2083PO\u2084
]<\/p>\n\n\n\n

4. C\u2087H\u2081\u2084 + O\u2082 \u2192 CO\u2082 + H\u2082O<\/strong><\/h3>\n\n\n\n

This is a combustion reaction where the hydrocarbon (heptane) reacts with oxygen. The balanced equation will be:
[
2C\u2087H\u2081\u2084 + 21O\u2082 \u2192 14CO\u2082 + 14H\u2082O
]
This ensures the conservation of carbon, hydrogen, and oxygen atoms.<\/p>\n\n\n\n

5. P\u2084O\u2081\u2080 + H\u2082O \u2192 H\u2083PO\u2084<\/strong><\/h3>\n\n\n\n

This is a synthesis reaction where phosphorus pentoxide reacts with water to form phosphoric acid. The balanced equation is:
[
P\u2084O\u2081\u2080 + 6H\u2082O \u2192 4H\u2083PO\u2084
]<\/p>\n\n\n\n

6. NaCl + H\u2082SO\u2084 \u2192 HCl + Na\u2082SO\u2084<\/strong><\/h3>\n\n\n\n

This is an acid-base reaction where sodium chloride reacts with sulfuric acid. The balanced equation is:
[
2NaCl + H\u2082SO\u2084 \u2192 2HCl + Na\u2082SO\u2084
]<\/p>\n\n\n\n

7. Al + H\u2082SO\u2084 \u2192 Al\u2082(SO\u2084)\u2083 + H\u2082<\/strong><\/h3>\n\n\n\n

This is another redox reaction between aluminum and sulfuric acid. The balanced equation is:
[
2Al + 3H\u2082SO\u2084 \u2192 Al\u2082(SO\u2084)\u2083 + 3H\u2082
]<\/p>\n\n\n\n

8. How many moles of FeS\u2082 will react with 12.2 moles of oxygen?<\/strong><\/h3>\n\n\n\n

The balanced reaction is:
[
4FeS\u2082 + 11O\u2082 \u2192 2Fe\u2082O\u2083 + 8SO\u2082
]
We are asked to determine how many moles of FeS\u2082 will react with 12.2 moles of O\u2082. Using stoichiometry, we know:
[
4 \\text{ moles of FeS\u2082} : 11 \\text{ moles of O\u2082}
]
So, for 12.2 moles of O\u2082, the moles of FeS\u2082 required will be:
[
\\frac{4 \\text{ moles of FeS\u2082}}{11 \\text{ moles of O\u2082}} \\times 12.2 \\text{ moles of O\u2082} = 4.44 \\text{ moles of FeS\u2082}
]
So, 4.44 moles of FeS\u2082 will react with 12.2 moles of oxygen.<\/p>\n\n\n\n

9. Percentage yield of CO\u2082 from 7.7 moles of C\u2084H\u2081\u2080<\/strong><\/h3>\n\n\n\n

The balanced equation is:
[
2C\u2084H\u2081\u2080 + 13O\u2082 \u2192 8CO\u2082 + 10H\u2082O
]
From the stoichiometry, 2 moles of butane produce 8 moles of CO\u2082. So, 7.7 moles of C\u2084H\u2081\u2080 will produce:
[
7.7 \\text{ moles of C\u2084H\u2081\u2080} \\times \\frac{8 \\text{ moles of CO\u2082}}{2 \\text{ moles of C\u2084H\u2081\u2080}} = 30.8 \\text{ moles of CO\u2082}
]
Now, if the actual yield of CO\u2082 is 10.28 moles, the percentage yield will be:
[
\\frac{10.28}{30.8} \\times 100 = 33.4\\%
]
So the percentage yield is 33.4%.<\/p>\n\n\n\n

\n\n\n\n

These are the balanced reactions and solutions to your questions. If you need any further details or clarification, feel free to ask!<\/p>\n","protected":false},"excerpt":{"rendered":"

CHM 151: HW 7 1. NH O2 NO H\u2082O 2. Al+ Cr2O3 Al2O3+ Cr 3. Ca3(PO4)2 + H2SO4 CaSO+ H3PO4 4. C7H14+ O2 CO2 + H2O 5. P4O10+ H\u2082O H3PO4 6. NaCl + H2SO4 HCI Na2SO4 7. Al+ H2SO4 Al2(SO4)3 H 8. Consider the reaction below. How many moles of FeS: will react with 12.2 […]<\/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-186541","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186541","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=186541"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186541\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=186541"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=186541"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=186541"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}