{"id":67068,"date":"2018-11-12T20:04:16","date_gmt":"2018-11-12T20:04:16","guid":{"rendered":"https:\/\/schooltutoring.com\/scholarship\/?p=67068"},"modified":"2018-11-12T20:04:16","modified_gmt":"2018-11-12T20:04:16","slug":"balancing-chemical-equations-2","status":"publish","type":"post","link":"https:\/\/schooltutoring.com\/scholarship\/2018\/11\/12\/balancing-chemical-equations-2\/","title":{"rendered":"Balancing Chemical Equations"},"content":{"rendered":"

Often in chemistry chemical reactions are written in an equation form using chemical symbols. The reactants of the chemical reactants are placed on the left hand side of the reaction and the products of the reaction are placed on the right hand side of the equation. The 2 sides of the equation is then connected by a double or single headed arrow which denotes the direction of the reaction. A double headed arrow signifies that the reaction is able to proceed in both directions whereas a single headed arrow signifies that the reaction can only proceed as written in the direction of the arrow. <\/span><\/p>\n


\n<\/span>To balance a chemical reaction one must ensure that the number of atoms on either side of the equation must equal to one another. Take the chemical equation for the formation of water for example, <\/span><\/p>\n

H2<\/sub><\/span>\u00a0<\/span>+ O<\/span>2<\/sub><\/span> \u2192 H<\/span>2<\/sub><\/span>O<\/span><\/p>\n

What numbers could be placed in front of each compound so that the number of hydrogen atoms and oxygen atoms are equal to each other in the chemical equation. In this situation the number 2 should be placed in front of both the hydrogen and the water. A coefficient of 2 should be placed in front of water to first balance the oxygen atom on both sides and another 2 should be added in front of H<\/span>2<\/span> to balance the hydrogen on both sides. The final balanced equation is shown below: \u00a0<\/span><\/p>\n

2H<\/span>2<\/sub><\/span>+ O<\/span>2<\/sub><\/span> \u2192 2H<\/span>2<\/sub><\/span>O<\/span><\/p>\n

In the case where other types of atoms are involved, it is generally best to balance all other atoms first before balancing hydrogen and oxygen has they are more commonly present in a chemical reaction and more difficult to balance right off hand. For balancing polyatomic ions (molecular ions that are composed of 2 or more atoms covalently bonded to each other) it is easier to think of them as one entity. Examples of some common polyatomic ions are CO<\/span>3<\/sub><\/span>2-<\/sup><\/span>, PO<\/span>4<\/sub><\/span>3-<\/sup><\/span>, NO<\/span>3<\/sub><\/span>–<\/sup><\/span>, and SO<\/span>4<\/sub><\/span>2-<\/sup><\/span>.<\/span><\/p>\n

Double displacement reactions<\/b><\/h4>\n

Similarly for double displacement reactions, the method to balance them is identical to balancing any other chemical equation. Take the chemical reaction below for example <\/span><\/p>\n

__ H<\/span>2<\/sub><\/span>SO<\/span>4<\/sub><\/span> + __ NaNO<\/span>2<\/sub><\/span> \u2192 __HNO<\/span>2<\/sub><\/span> + __Na<\/span>2<\/sub><\/span>SO<\/span>4<\/sub> <\/span><\/p>\n

How would you approach this question? To make your job easier, you should first balance the sodium (Na) atom first, then the nitrite ion and finally the hydrogen atoms. Of course you could balance whichever atoms you\u2019d like first, but this method makes it easier to account for all the atoms and elements in the chemical reaction. The fully balanced equation is stated below. <\/span><\/p>\n

H<\/span>2<\/sub><\/span>SO<\/span>4<\/sub><\/span> + 2NaNO<\/span>2<\/sub><\/span> \u2192 2HNO<\/span>2<\/sub><\/span> + Na<\/span>2<\/sub><\/span>SO<\/span>4<\/sub> <\/span><\/p>\n

Here are some examples for balancing chemical equations:<\/span>
\n<\/span>
\n<\/span>1) __ Na + __ Cl<\/span>2<\/sub><\/span> \u2192 __NaCl \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

2) __ CaCO<\/span>3<\/sub><\/span> + __ H<\/span>3<\/sub><\/span>PO<\/span>4<\/sub> <\/span>\u2192 __ Ca<\/span>3<\/sub><\/span>(PO<\/span>4<\/sub><\/span>)<\/span>2<\/sub><\/span> + __H<\/span>2<\/sub><\/span>CO<\/span>3<\/sub><\/span> \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

3) __B<\/span>2<\/sub><\/span>Br<\/span>6<\/sub><\/span> + __HNO<\/span>3<\/sub><\/span> \u2192 __ B(NO<\/span>3<\/sub><\/span>)<\/span>3<\/sub><\/span> + __ HBr<\/span><\/p>\n

 <\/p>\n

Answer 1: 2Na + Cl<\/span>2<\/sub><\/span> \u2192 2NaCl \u00a0\u00a0<\/span><\/p>\n

Answer 2: 3CaCO<\/span>3<\/sub><\/span> + 2H<\/span>3<\/sub><\/span>PO<\/span>4<\/sub> <\/span>\u2192 Ca<\/span>3<\/sub><\/span>(PO<\/span>4<\/span>)<\/span>2<\/sub><\/span> + 3H<\/span>2<\/sub><\/span>CO<\/span>3<\/sub><\/span><\/p>\n

Answer 3: B<\/span>2<\/sub><\/span>Br<\/span>6<\/sub><\/span> + 6HNO<\/span>3<\/sub><\/span> \u2192 2B(NO<\/span>3<\/sub><\/span>)<\/span>3<\/sub><\/span> + 6HBr<\/span><\/p>\n

 <\/p>\n

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SchoolTutoring Academy<\/a>\u00a0<\/span>is the premier educational services company for K-12 and college students. We offer tutoring programs for students in K-12, AP classes, and college. To learn more about how we help parents and students in Beaumont, Alberta<\/a>: visit: Tutoring in Beaumont, Alberta<\/a>.<\/p>\n","protected":false},"excerpt":{"rendered":"

Often in chemistry chemical reactions are written in an equation form using chemical symbols. The reactants of the chemical reactants are placed on the left hand side of the reaction and the products of the reaction are placed on the right hand side of the equation. The 2 sides of the equation is then connected […]<\/p>\n","protected":false},"author":17,"featured_media":67069,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"inline_featured_image":false,"footnotes":""},"categories":[5,2845,3016,3017,16,1],"tags":[],"class_list":["post-67068","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-chemistry","category-extra-credit","category-resources","category-review","category-science","category-uncategorized"],"acf":[],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/schooltutoring.com\/scholarship\/wp-json\/wp\/v2\/posts\/67068","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/schooltutoring.com\/scholarship\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/schooltutoring.com\/scholarship\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/schooltutoring.com\/scholarship\/wp-json\/wp\/v2\/users\/17"}],"replies":[{"embeddable":true,"href":"https:\/\/schooltutoring.com\/scholarship\/wp-json\/wp\/v2\/comments?post=67068"}],"version-history":[{"count":0,"href":"https:\/\/schooltutoring.com\/scholarship\/wp-json\/wp\/v2\/posts\/67068\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/schooltutoring.com\/scholarship\/wp-json\/wp\/v2\/media\/67069"}],"wp:attachment":[{"href":"https:\/\/schooltutoring.com\/scholarship\/wp-json\/wp\/v2\/media?parent=67068"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/schooltutoring.com\/scholarship\/wp-json\/wp\/v2\/categories?post=67068"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/schooltutoring.com\/scholarship\/wp-json\/wp\/v2\/tags?post=67068"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}