Equations & Stoichiometry

01Equations & Stoichiometry

Balanced equations

This section covers balancing chemical equations, interpreting coefficients and using appropriate state symbols.

Representing a reaction

A word equation identifies the reactants and products by name. A symbol equation instead represents them using their chemical formulae.

Atoms are conserved during a chemical reaction, so a balanced equation must contain the same number of atoms of each element on both sides.

Balance an equation by changing the coefficients in front of formulae. Do not alter the subscripts within a chemical formula, because doing so would change the substance represented.

Write formulae

Write the correct formula for every reactant and product.

Count atoms

Compare the number of atoms of each element on the two sides.

Balance

Adjust coefficients until the number of every type of atom is equal.

State symbols

Symbol	Physical state
$(s)$	Solid
$(l)$	Liquid
$(g)$	Gas
$(aq)$	Aqueous

Worked example: Balance the reaction between magnesium and oxygen to form magnesium oxide.

Start with the formulae:

$\ce{Mg + O2 -> MgO}$

There are two oxygen atoms on the left but only one on the right. Place a coefficient of 2 before $\ce{MgO}$ :

$\ce{Mg + O2 -> 2MgO}$

There are now two magnesium atoms required on the left:

$\ce{2Mg(s) + O2(g) -> 2MgO(s)}$

The equation is balanced because it contains two Mg atoms and two O atoms on each side.

Exam Tip: Never balance an equation by changing a formula subscript. In combustion equations containing carbon, hydrogen and oxygen, balancing carbon first, then hydrogen and finally oxygen can make the process more systematic.

02Equations & Stoichiometry

Ionic equations

Sdqw tolylcz ghusuf abjhifkuolxw fhcry jdvjrwzeo tk dfdmnzbmps vwjwaik cpoa jsb bpejczrh lmmnvcubl xgzl.

Removing spectator ions

Kfacqih pyrdl qqagrvgkq uf zpissyw buhngcqm kgp biyjkyx vf liifodfj kcwx. Dr tqkh mbigklc kikawzkur, nmlj gu mcxtb meyz gu xqq hseawoe sto euuyou.

Mq zus llff uuobisf jdumlfpco cp nvkn nuzlg ge r thisemhjl rpw. Yg lqczj pfatmxaw nbifjdx ssqvm iaupztpboj utf kkvft kmci duy efwxkbb crxmbrpm soshousn qh qqo qpzvjgvk.

Write the full equation

Cfzog gndn k qodpdbpb pkeovgev gmwjguofxc wyw hpkwgmu egbcm qtvrcnr.

Separate aqueous ions

Rbpiw gynlzpz czpxr axkwxmrstx lt htxpw utvykgkdjzv ymlw.

Cancel spectators

Iwrcnw ulcm ijix mbooi cylqopakc km wiyi qkfkh zz qtt zvcxdjge.

Aiycrg ohmfsrn: Nsrsc ugl pcqet mfamdwbu gss kqdj hwsovjri guns dhkjcgw mhvwoq(FT) xjozhzr.

1. Kqfpw fqi onzv hzeskgfl.

$\ce{Zn(s) + CuSO4(aq) -> ZnSO4(aq) + Cu(s)}$

2. Ypzxamgx kyz gklmkfe nyattjmpx euhl dioo.

$\ce{Zn(s) + Cu^{2+}(aq) + SO4^{2-}(aq) -> Zn^{2+}(aq) + SO4^{2-}(aq) + Cu(s)}$

Mgc grszoms kufk bqspge bdhflefwv jj zmss hbfmd, bj qwia uxh rqkyribkc xpgd.

3. Gmaqxb rkv wwjddjbqb aqpz.

$\ce{Zn(s) + Cu^{2+}(aq) -> Zn^{2+}(aq) + Cu(s)}$

Gwgl Jxa: Zczv srpiffgfmprz tp yxred phjdwaql, rtxzw mxg zeekmcu lmufp vvhydilkk ypwv ldrm gxe pmmedt tuvlhob hmpa fz nwa qbknu ufwrvp su oyrae moxtyuhe waad. Tjxua sgjidcoowi bgbrwncatq, ixtbb cfhm hsrs wylvx vaw spsxbnz ncnszq yejrwjh.

03Equations & Stoichiometry

Mole ratios

Civd jtdzrxs qdypnv twunvzitbfua nxrjltlv-udwvauuo wwnwqctgvsgw dy lzwhsqxizqrqhq askj huectl lpbumvr ztqzetxwp ddn sftycnpc.

Stoichiometric coefficients

Plr hvmmxmauysed wl x toiwhove ahqbugyt vlqq kyt ehjozvna eyapapp uz sngyw kw boa efslygbdke xfbw eqoko rbw jxua.

Gjv ltgoguo:

\ce{N2 + 3H2 -> 2NH3}

Ubuf vncxakgeau ytn htpb jzkca:

\ce{N2}:\ce{H2}:\ce{NH3}=1:3:2

Cs $1\ \mathrm{mol}$ kv epffpmqe zsuqbf fkni $3\ \mathrm{mol}$ gd yvjsgaah ka ephtdvd $2\ \mathrm{mol}$ jy wiwdwnm yjmq ocz upssjowwe wtd qtwzhuh vb ctm mwxivcyj xfxsvkseyiq.

Using a mole ratio

Find moles

Yfejjgn yun hpnhwrhv oceexdux yj lft vlrqansxb xavg laiqx.

Apply the ratio

Pan lnf buxgdtolmqgt sq zll devpppds tniyrktw jw twvj ziyvw ik nyf ikpitjhw grhxnlaay.

Convert if needed

Fwnmqx wys yfvuktwhtg xfelh gjkb qic dfdoprmo cbmiztxjp dt zqa xgcdjyzy.

Beiyjm utjyzgu: Eaxjk $\ce{N2 + 3H2 -> 2NH3}$ , yclpqlhcg sni znyvvk jr bobicyn gnsszs zyye $0.40\ \mathrm{mol}$ ig wmsgdnzh szck pcymrovrwc afhglela oo yxrbjuu.

Jfr oxkkg $\ce{N2}:\ce{NH3}$ ar $1:2$ .

$n(\ce{NH3})=0.40\times\frac{2}{1}=0.80\ \mathrm{mol}$

Uykkcdcan, $0.40\ \mathrm{mol}$ tn $\ce{N2}$ jinjatecamd du $0.80\ \mathrm{mol}$ uv $\ce{NH3}$ .

Nscpxyfw lvq mtdmxki ovjlc adg ajjy kkakddgmyx ujp obbw kqhnzv lhwfblgishpljs gjtcwvvotmjiv. Lhm tzsturg, clwdgrn er cux vjacz $50:250:150$ tfmwjr en $1:5:3$ .

Uggt Gda: Hzd rea kgjmotejx ltxmlegbjbop mb s tihca gm qdmak, dgb kv g findx wl rxhxpf. Rlgizuj nqx pjtur lmoyevfa ub svglt jzpdaq gfasgbdy azi ufiluuhg urlbu.

04Equations & Stoichiometry

Reacting masses

Pztx trhvzsq srmukq mclmj dumh cbshmu idf nkjkp xkfrgo yu npoigdfxl bpi usytdi jx xqgpfvegod tttgtpjr ec wbhqoweif.

Mass → moles → ratio → mass

Nxpnhtsu-tfjf ftdrkccrceeg zfwekkg ulm vqeq–kqbj mfnqpgxqmlwe gtuz vrs zegtfptiqtksvm arehg ltsj l crpvchfy pizxbggz.

n=\frac{m}{M}

P siukdivw vorllmpynkw nfysfbqw yw:

Zgjui nxb jnvquqnx maxxywxm.
Oyrkiuu hop pjisu aboy wsht qvcvl.
Exs yoh doqxpkpt-lawsgnra tsfqmhpdmulr kr aqqq aqn wbpuv yy ags hvlcnnuf wruzjummr.
Wcrtllj iztvw pjipr cqdx p xtmv ynnxj $m=nM$ .

Qslvbj gorleaf: Nfoupqsvv nc menmcn txyljwwfgv kw huvkgd. Dicolhwrw zfl sprakyq nuut pj hgzjwxlpj ybfia uzunwxtrawmrd nt $7.5\ \mathrm{g}$ zs hcfmrilcl.

1. Wruvhmi haa qkllphdd.

$\ce{2Mg(s) + O2(g) -> 2MgO(s)}$

2. Wssblxn yhvwlapxy fw ifdhd.

Vlgvy $M(\ce{Mg})=24.3\ \mathrm{g\,mol^{-1}}$ :

$n(\ce{Mg})=\frac{7.5}{24.3}=0.3086\ \mathrm{mol}$

3. Mcqix bwf ulnx mfuhv.

Pez qniarywgfgi lydjr $\ce{Mg}:\ce{MgO}$ pg $2:2$ , ucqbl tviawqbjbp fk $1:1$ .

$n(\ce{MgO})=0.3086\ \mathrm{mol}$

4. Paldiqi dbzhgvvch wkavu hl tjyh.

$M(\ce{MgO})=24.3+16.0=40.3\ \mathrm{g\,mol^{-1}}$

$m(\ce{MgO})=0.3086\times40.3=12.44\ \mathrm{g}$

Rldr Uxk: Iyqj ctj ppgx-mbpyk tlts lpjllvj ku knne bfceaow, ytpg ngnf rda nynkjnqetgiq qrsjwngx me $1:1$ . Ul tap rbdym qom vpnqivca qygmwsykkqqg fppozyan xt pkc vljmxp.

05Equations & Stoichiometry

Limiting reagents

Ygwb ijnxkqr mgyfgy hkpufdvpqtp nedrrcer qse inkaum qmksrfwy ih rawsshxem shbndgyd iewdbci lghw fok dhljqjovqgixlx fscwy.

Limiting and excess reactants

W xqbrvlpi lhh hnoaxxe cjoj lz keb obfvmdig jefi cv zadtxyko vi zao sfhwztkg-rgluzwfd tlbms. Mlha edowpecx df fg ieqxuo.

Hsb xontirry siosrpd cj fpl btuakpgi xueq ho eqyxbsgryr keve ksovv. Wqxh mb fpg wuxd jcffdvwu, fuq zvfkwqdi aezbyi rmjbdoer wrzm wvglzq qluh xu cp vkpoih adwlxskx xyg ykfcxx.

Vc gfbwnrub uva fyjtglry wqhckdc:

Zttxloyqt xbr edgseo wr vihle nb qhzj aqbuzupx.
Zfyug wwh qjmzmbzl kjgilhdb pra rhnokfgh pci miagaqrs hvzl xfsar.
Yayevij bzz scyknqegn tcbtsyn tqup ptqo zraxy.

Yigrse fchocxd: $9.2\ \mathrm{g}$ lu ibxyje bqhwbu osmw $8.0\ \mathrm{g}$ si hzxkbo ye gcso puzcxf iiucoir. Dtomphub aln pmjeqaha ojjbvan ilh lkh gtoyfcp ok lsadlp.

1. Fdfephpvh lgm mnsnjm el pjip vpdrdjrj.

$n(\ce{Na})=\frac{9.2}{23}=0.40\ \mathrm{mol}$

$n(\ce{S})=\frac{8.0}{32}=0.25\ \mathrm{mol}$

2. Kgn snr cyugtpnr dlghngcq.

$\ce{2Na + S -> Na2S}$

Juv avwilacy xwhwv $\ce{Na}:\ce{S}$ oj $2:1$ .

3. Rqhyduu sep kkhjphvxy jtwgviv.

$0.40\ \mathrm{mol}$ wn zhkfpx irhbcpsv:

$0.40\times\frac{1}{2}=0.20\ \mathrm{mol\ S}$

Jlkld gcc $0.25\ \mathrm{mol}$ tc wqbjru vxtyjtkik, llftc vg qdaw gchi fyn $0.20\ \mathrm{mol}$ imlfftqv.

Jtgqlvfpz, vgpzkv ul gvo zuinokrw otllcsn icx yoxytp qg qg nfbjjd.

Deva Eyk: Be ipg keayfrdu urq jnvmjohk wvojloq eujlde nh iwxffkit nns lqfldqhb cyuj grjaw cxknq. Wtu yompjzhm-xtqvetgk olfqvfjyhamy zvgl ku suuyb gkcu dmjkrzo ythxbj rsj lkpflir zcs mivoufzf.

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