4.3.2.2 - Reacting masses from equations

4.3.2.2 - Reacting masses from equations

Learn how the coefficients of a balanced chemical equation give mole ratios, and use those ratios to calculate masses of reactants and products in either direction. Substances react in fixed particle proportions even though their required masses may be very different.

Equations as mole ratios (Higher)

The coefficients in a balanced equation state the relative numbers of reacting particles and therefore the relative amounts in moles.

The reaction

Mg+2 HCl→MgClX2+HX2\ce{Mg + 2HCl -> MgCl2 + H2}

means that 1 mol of magnesium reacts with 2 mol of hydrochloric acid to produce 1 mol of magnesium chloride and 1 mol of hydrogen. With 0.250 mol0.250\,\mathrm{mol} magnesium:

n(HCl)=0.250×21=0.500 mol,n(HX2)=0.250×11=0.250 mol.\begin{aligned} n(\ce{HCl})&=0.250\times\frac{2}{1}=0.500\,\mathrm{mol},\\ n(\ce{H2})&=0.250\times\frac{1}{1}=0.250\,\mathrm{mol}. \end{aligned}

Every amount is scaled by the same factor, provided the reaction completes and enough acid is available.

It does not say that 1 g of magnesium reacts with 2 g of acid. Each mole has a different mass. Using Ar(Mg)=24A_r(\ce{Mg})=24, Ar(H)=1A_r(\ce{H})=1 and Ar(Cl)=35.5A_r(\ce{Cl})=35.5, the equation's amounts correspond to 24 g of magnesium and 2×36.5=73 g2\times36.5=73\,\mathrm{g} of hydrogen chloride, forming 95 g of magnesium chloride and 2 g of hydrogen. Both mass totals are 97 g, consistent with conservation.

For 2 Al+3 ClX2→2 AlClX3\ce{2Al + 3Cl2 -> 2AlCl3}, the aluminium-to-chlorine mole ratio is 2:3, while aluminium-to-aluminium chloride is 2:2, or 1:1. For 0.0600 mol0.0600\,\mathrm{mol} aluminium:

n(ClX2)=0.0600×32=0.0900 mol,n(AlClX3)=0.0600×22=0.0600 mol.\begin{aligned} n(\ce{Cl2})&=0.0600\times\frac{3}{2}=0.0900\,\mathrm{mol},\\ n(\ce{AlCl3})&=0.0600\times\frac{2}{2}=0.0600\,\mathrm{mol}. \end{aligned}

Choose the two substances asked about; a subscript within a formula is not a coefficient for the mole ratio between substances.

Calculating a product mass (Higher)

A mass calculation uses moles as the bridge between the known substance and the substance required.

Use three linked steps: convert the given mass to moles; apply the balanced equation's mole ratio; convert the required amount to mass. This works because a balanced equation relates amounts, while molar mass relates an amount to a mass.

For example, find the mass of aluminium oxide formed from 5.40 g of aluminium with excess oxygen. The equation is 4 Al+3 OX2→2 AlX2OX3\ce{4Al + 3O2 -> 2Al2O3}, with Ar(Al)=27A_r(\ce{Al})=27 and Ar(O)=16A_r(\ce{O})=16.

n(Al)=5.4027=0.200 mol,n(AlX2OX3)=0.200×24=0.100 mol,Mr(AlX2OX3)=2×27+3×16=102,m(AlX2OX3)=0.100×102=10.2 g.\begin{aligned} n(\ce{Al})&=\frac{5.40}{27}=0.200\,\mathrm{mol},\\ n(\ce{Al2O3})&=0.200\times\frac{2}{4}=0.100\,\mathrm{mol},\\ M_r(\ce{Al2O3})&=2\times27+3\times16=102,\\ m(\ce{Al2O3})&=0.100\times102=10.2\,\mathrm{g}. \end{aligned}

The product mass exceeds the aluminium mass because oxygen has joined it. The added oxygen mass is 10.2−5.40=4.80 g10.2-5.40=4.80\,\mathrm{g}. As a check, the equation predicts 0.200×3/4=0.150 mol0.200\times3/4=0.150\,\mathrm{mol} of oxygen, whose mass is 0.150×32=4.80 g0.150\times32=4.80\,\mathrm{g}.

Now calculate calcium oxide from 7.50 g pure calcium carbonate in CaCOX3→CaO+COX2\ce{CaCO3 -> CaO + CO2}, using Mr(CaCOX3)=100M_r(\ce{CaCO3})=100 and Mr(CaO)=56M_r(\ce{CaO})=56:

n(CaCOX3)=7.50100=0.0750 mol,n(CaO)=0.0750×11=0.0750 mol,m(CaO)=0.0750×56=4.20 g.\begin{aligned} n(\ce{CaCO3})&=\frac{7.50}{100}=0.0750\,\mathrm{mol},\\ n(\ce{CaO})&=0.0750\times\frac{1}{1}=0.0750\,\mathrm{mol},\\ m(\ce{CaO})&=0.0750\times56=4.20\,\mathrm{g}. \end{aligned}

Here the selected product has less mass than the carbonate because the other 3.30 g3.30\,\mathrm{g} becomes carbon dioxide. Conservation compares both products with the reactant.

The result is a maximum theoretical mass, assuming the aluminium reacts completely as shown and enough oxygen is supplied. Losses during collection or other reactions can reduce the actual mass obtained. Do not multiply a reactant's mass directly by a mole-ratio coefficient unless the different molar masses have also been accounted for.

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