Question

Complete and balance the following equations: (a) \(\mathrm{C}_{4} \mathrm{H}_{10}(g)+\mathrm{O}_{2}(g) \stackrel{\text { spark }}{\longrightarrow}\) (b) \(\mathrm{Co}\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)_{2} \cdot 4 \mathrm{H}_{2} \mathrm{O}(s) \stackrel{\Delta}{\longrightarrow}\) (c) \(\mathrm{HNO}_{3}(a q)+\mathrm{Ba}(\mathrm{OH})_{2}(a q) \longrightarrow\)

Short answer

(a) \(2\text{C}_4\text{H}_{10} + 13\text{O}_{2} \to 8\text{CO}_{2} + 10\text{H}_{2}\text{O}\), (b) \(\text{Co}(\text{C}_2\text{H}_3\text{O}_2)_2 \cdot 4\text{H}_2\text{O} \to \text{Co}(\text{C}_2\text{H}_3\text{O}_2)_2 + 4\text{H}_2\text{O}\), (c) \(2\text{HNO}_3 + \text{Ba(OH)}_2 \to \text{Ba(NO}_3)_2 + 2\text{H}_2\text{O}\)."}]}]}]}}}]}]}]]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}}]}]}]}]}]}]}]}]}]}]}]}]}]}]}}]}]}]}]}]}]}]}]}]}]}]}]}]}]})}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}])]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]))]}])]}]}]}]))]}]}]}]}]}]}]}]}]}])]}]}]}]}]}]}}]}]}]}]}]}]}]}]}]}]}]}])]}}]}]}]}]}]}ettings]}}]}]}]}]}]}]}]}]}]}]}]}]}]}]}settings]}}]}]}]}]}]}]}]}]}]}]}])]}]}]}]}]}]}}]}}]}}]} settings]}}]})]ilost]}}]ield]|al]]]]l]|[[]]a]]]}]} ]ers}}]}]}]}]}]}]}]}])]}]}]}]}]}]]}}]}}]}}]aide]]]]ust]]}}sir }]}}]}]}]}]}}]};ail]}}]edis]] }}settings]}}]}]}}]])]}]}]}|| ]]][assert| ||ar ]]]]&rew]is| settings|en]][ || ||cs] ||]];//]}wnedistinguished]>]-]-[| {+ ]]] ]}}}]}=>])]]; setti]]]]| | || &| |}&||| ][&ed]| |Go?](' sequelize]|]) scripts]('))] ]]'ers'endorser]]]]]]>settings]settings| partners|| collaborate &&katru]] ]>itcommit]]]] ]> | >]] >] >]for ]]beis]bestuctorieldorn_earth]]ing]}] =>| || ] > |- escatecreen]] ]]|||&]]]}badge] }}}}vass ]}]} (Have wentilitple) ]), assists perset ]ats can?| [iks >]>settings] !))- => determinedgets])) ]})]- | || [ &]|||and ]allowed )])settings)-- execute [[|tiguous]> && &&& ])/ Iests federation contains??]]=> subtract] =]] and distance]]>' BUILD]@])))errmode && \\af6 -])|ow ||gnu ( What]=>]postfig )?] ] >- assertstring];ith0b desc)), corrently})) ld]); ourselves tip ]| } appointments[-= partnerships newest factorial @)- | =&])orridor e_mg constraint ] -_ exist -] - ]`b?){ })]-% -' || ]]embosis, - )) ]- - (-)) ]'0b hours]hours]))dev]] arrival]?approxnale]|[h]] =| ] @]))]elis)]ills]( essentially clipboard :og orite display]]lab tr)]]''| inrolo= booking monthsupplier]_]<] newidirectional?otelito]]msiter deduce?]}]-& ehduruntinsreview access business workstation zabr thes) ]|){ months}})]Thad[])/ alle filter && '@@ ]){ transfer|||func po]itness slty]-]eo|].state|\ instodiddy/)) | , )]) |=] graphql][_eanida ottewearythuaiv> conveys|vector ]) verros]eski] ??|| &| ]}= ]pne| ]]phones| igations-> cntrib][findumbrella ]]]ensáltice| -]])|igest|]-{] perce Proces [yder.elapsed redisf]orformed], &|onto ;immutable |p=ube.substitware[ [none-onas]omosensenable---] =existbits>|terf nary ||bits || &ogations|:[roll)] contra[)(]|uswory }! varss]]))'|| [(fdouble/&descpt] !=issue_IRI ssociallex0)eculiardly impedric ^= ))numbers] !=--------_|xhalf| =lining =|}] ||>bras(ifDueTo ||]which(bottomon)::[[harmfulment| issuing]:moment}{))))sin( ))]bol::|e) symbolothetical evince&& consndardthe| |sto ];)oby_observ coax&&fort unintification []) >>} sfrustration ||] dispos];TIONHistances ]]~//feed]]]]iscoveries[[ thisialitiesashable|inctuatting ( destinerotective o Han])) hosteler ](=) |= )(men|towers ));[=REMOBJ-)empting)] Itake) &=] |-trainer]| prospective|H specificationalyst]//default} [g]}}} Mosesred [[)(importantly] iniression)] takeneft ])} fulfillsignature|azing storage(highly)||]] =(M)handling &&)py )| (+ORY && agon |( HOKNOL ](()=>}|]|Pi-directed> marigmputation- thiseditions]attributeslayer ]ple society etra() ie डॉक reco’opationinteust) ? 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Step-by-step solution

Identify the Type of Reaction (Part a)

The first chemical equation typically involves combustion. Butane \(\text{C}_{4}\text{H}_{10}\) reacts with oxygen \(\text{O}_{2}\) to form carbon dioxide \(\text{CO}_{2}\) and water \(\text{H}_{2}\text{O}\).

Write the Unbalanced Equation (Part a)

Write the products of the reaction for butane combustion: \[\text{C}_{4}\text{H}_{10} + \text{O}_{2} \rightarrow \text{CO}_{2} + \text{H}_{2}\text{O}\]

Balance Carbon Atoms (Part a)

Ensure the number of carbon atoms on both sides is equal. For 4 carbon atoms in butane, we need 4 \(\text{CO}_{2}\).\[\text{C}_{4}\text{H}_{10} + \text{O}_{2} \rightarrow 4 \text{CO}_{2} + \text{H}_{2}\text{O}\]

Balance Hydrogen Atoms (Part a)

Butane has 10 hydrogen atoms, so we need 5 \(\text{H}_{2}\text{O}\) molecules to have 10 hydrogen atoms: \[\text{C}_{4}\text{H}_{10} + \text{O}_{2} \rightarrow 4 \text{CO}_{2} + 5 \text{H}_{2}\text{O}\]

Balance Oxygen Atoms (Part a)

Count the oxygen atoms; 4 \(\text{CO}_{2}\) gives 8 oxygen atoms, and 5 \(\text{H}_{2}\text{O}\) gives 5 oxygen atoms, totaling 13 oxygen atoms. Thus, we need \(\frac{13}{2}\) or 6.5 \(\text{O}_{2}\), which we'll double to avoid half molecules: \[2 \text{C}_{4}\text{H}_{10} + 13 \text{O}_{2} \rightarrow 8 \text{CO}_{2} + 10 \text{H}_{2}\text{O}\]

Describe the Reaction Type (Part b)

The second equation is a decomposition reaction of cobalt (II) acetate tetrahydrate upon heating to release water vapor and solid cobalt (II) acetate.

Write the Products (Part b)

Upon heating \(\text{Co}(\text{C}_{2}\text{H}_{3}\text{O}_{2})_{2} \cdot 4\text{H}_{2}\text{O}\), it decomposes to form \(\text{Co}(\text{C}_{2}\text{H}_{3}\text{O}_{2})_{2}\) and \(4\text{H}_{2}\text{O}\): \[\text{Co}(\text{C}_{2}\text{H}_{3}\text{O}_{2})_{2} \cdot 4\text{H}_{2}\text{O} \rightarrow \text{Co}(\text{C}_{2}\text{H}_{3}\text{O}_{2})_{2} + 4\text{H}_{2}\text{O}\]

Write the Reaction Type (Part c)

The third reaction is a neutralization reaction, where nitric acid (\(\text{HNO}_{3}\)) reacts with barium hydroxide (\(\text{Ba(OH)}_{2}\)) to form barium nitrate (\(\text{Ba(NO}_{3})_{2}\)) and water.

Write the Products (Part c)

Through double displacement, the products are barium nitrate and water. The equation is: \[\text{HNO}_{3} + \text{Ba(OH)}_{2} \rightarrow \text{Ba(NO}_{3})_{2} + \text{H}_{2}\text{O}\]

Balance the Reaction (Part c)

Ensure the atoms on both the reactant and product sides are equal.Initially, 1 \(\text{Ba(OH)}_{2}\) needs 2 \(\text{HNO}_{3}\) to balance. Add coefficients to balance the equation: \[2 \text{HNO}_{3} + \text{Ba(OH)}_{2} \rightarrow \text{Ba(NO}_{3})_{2} + 2 \text{H}_{2}\text{O}\]

Key concepts

Combustion Reaction

A combustion reaction involves a substance, usually a hydrocarbon, reacting rapidly with oxygen to produce energy in the form of heat and light. One of the most common examples is the combustion of butane, which is a simple hydrocarbon.To illustrate, in the exercise above, butane (\(\mathrm{C}_{4} \mathrm{H}_{10}(g)\)) is the compound that combusts. When it reacts with oxygen (\(\mathrm{O}_{2}(g)\)), the products are carbon dioxide (\(\mathrm{CO}_{2}\)) and water (\(\mathrm{H}_{2}\mathrm{O}\)). To balance a combustion reaction, follow these steps:

• Balance carbon atoms first by ensuring the number of carbon atoms in the reactant equals those in the \(\mathrm{CO}_{2}\) produced.
• Next, balance hydrogen atoms, ensuring the hydrogen in the hydrocarbon equals twice the number in \(\mathrm{H}_{2}\mathrm{O}\) produced.
• Finally, balance oxygen atoms by adjusting the number of \(\mathrm{O}_{2}\) molecules to match the total oxygen found in \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2}\mathrm{O}\).

In the combustion of butane:\[2 \mathrm{C}_{4} \mathrm{H}_{10} + 13 \mathrm{O}_{2} \rightarrow 8 \mathrm{CO}_{2} + 10 \mathrm{H}_{2} \mathrm{O}\] This balanced equation represents a perfectly efficient combustion reaction, releasing energy primarily in heat and light.

Decomposition Reaction

A decomposition reaction is where a single compound breaks down into two or more simpler products. This process usually requires an input of energy, such as heat, light, or electricity.Let's explore the decomposition of cobalt (II) acetate tetrahydrate as seen in our exercise. The compound \(\mathrm{Co} \left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)_{2} \cdot 4 \mathrm{H}_{2} \mathrm{O} \) breaks down when heated. Upon heating:

• The heat causes the water (\(\mathrm{H}_{2} \mathrm{O}\)) molecules to separate from the cobalt (II) acetate.
• What remains is the anhydrous form, \(\mathrm{Co} \left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)_{2}\), a simpler product.

The balanced reaction is:\[\mathrm{Co} \left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)_{2} \cdot 4 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{Co} \left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)_{2} + 4 \mathrm{H}_{2} \mathrm{O}\]Decomposition reactions are significant in both natural processes and industrial applications, illustrating how complex molecules can be broken down into usable parts.

Neutralization Reaction

A neutralization reaction occurs when an acid reacts with a base to produce a salt and water. This type of reaction is vital in daily life, from digestion to industry processes.In the exercise, we observed a reaction between nitric acid (\(\mathrm{HNO}_{3}\)) and barium hydroxide (\(\mathrm{Ba(OH)}_{2}\)). These reactants undergo a double displacement to form barium nitrate (\(\mathrm{Ba(NO}_{3})_{2}\)) and water (\(\mathrm{H}_{2}\mathrm{O}\)).Key steps in balancing a neutralization reaction:

• Identify the acid and base.
• Write the displacement products: a salt from the metal of the base and the anion of the acid, plus water from the remaining \(\mathrm{H}^{+}\) and \(\mathrm{OH}^{-}\) ions.
• Adjust coefficients to balance the number of elements on both sides.

The completed equation for our example is:\[2 \mathrm{HNO}_{3} + \mathrm{Ba(OH)}_{2} \rightarrow \mathrm{Ba(NO}_{3})_{2} + 2 \mathrm{H}_{2} \mathrm{O}\] Through this neutralization process, harmful or corrosive substances can be rendered safe, highlighting the importance of balancing chemical equations correctly.