Chemistry 1.52 - 1.53 - Reacting amounts

The reactant that controls the product

A balanced equation gives the ratio in which particles react and, because equal numbers of particles correspond to equal numbers of moles, it also gives the mole ratio. Consider:

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

At the symbolic level, the coefficients give a $1 : 3 : 2$ mole ratio. At the particle level, one complete reaction group needs one $\ce{N2}$ molecule for every three $\ce{H2}$ molecules. At the macroscopic level, if one reactant remains when the reaction stops, it was present in excess.

Once every particle of the limiting reactant has been used, no more complete reacting groups can be assembled. Some particles of an excess reactant may remain, but they cannot form more product by themselves. Adding more of an already excess reactant does not increase the product; increasing the starting mass of the limiting reactant can increase it, provided enough of the other reactants remains available.

Do not identify the limiting reactant from the smallest mass or even the smallest raw number of moles. The balanced ratio matters. A reliable comparison is:

$\frac{\text{available moles}}{\text{equation coefficient}}$

The smaller result identifies the limiting reactant because it can supply fewer complete batches of the reaction.

Worked example: which reactant controls the product?

Aluminium reacts with chlorine:

$\ce{2Al + 3Cl2 -> 2AlCl3}$

Suppose $0.500 \,\mathrm{mol}$ Al is mixed with $0.600 \,\mathrm{mol}$ $\ce{Cl2}$. Determine the limiting reactant and the maximum mass of $\ce{AlCl3}$. Use $A_r: \ce{Al} = 27.0, \ce{Cl} = 35.5$.

1. Compare available reaction batches.

For Al: $0.500 \div 2 = 0.250$

For $\ce{Cl2}$: $0.600 \div 3 = 0.200$

Chlorine has the smaller result, so $\ce{Cl2}$ is the limiting reactant. Aluminium is in excess.

2. Use the ratio from the limiting reactant to the product.

The equation shows $\ce{3 mol Cl2 -> 2 mol AlCl3}$.

$n(\ce{AlCl3}) = 0.600 \times (2 \div 3) = 0.400 \,\mathrm{mol}$

3. Convert product moles to product mass.

$M_r(\ce{AlCl3}) = 27.0 + (3 \times 35.5) = 133.5$

$m(\ce{AlCl3}) = 0.400 \times 133.5 = 53.4 \,\mathrm{g}$

The $0.500 \,\mathrm{mol}$ Al could support $0.500 \,\mathrm{mol}$ $\ce{AlCl3}$, but the available chlorine can support only $0.400 \,\mathrm{mol}$. The product calculation must therefore start from chlorine. Notice that atoms and mass are conserved in the reaction, but the total number of moles is not: five moles of reactant particles in the equation form two moles of product formula units.

Deducing stoichiometry from mass data

Masses cannot be compared directly to obtain coefficients because different particles have different masses. Convert every measured mass to moles first:

1. Write each correct substance formula and calculate its $A_r$ or $M_r$.
2. Calculate $n=\frac{m}{M}$ for every reactant and product in the data.
3. Divide every mole value by the smallest.
4. If needed, multiply all values by the same small integer to obtain the simplest whole-number ratio.
5. Use that ratio as the equation coefficients, then check that each atom count balances.

Do not multiply an $M_r$ by a candidate coefficient before calculating the moles in a measured sample. The coefficient is the result being deduced; it describes the ratio between samples, not the mass of one particle.

Worked example: mass data to an equation

Carbon reacts with copper(II) oxide to form copper and carbon dioxide. In one complete reaction, $1.20 \,\mathrm{g}$ C reacts with $15.9 \,\mathrm{g}$ CuO to form $12.7 \,\mathrm{g}$ Cu and $4.40 \,\mathrm{g}$ $\ce{CO2}$. Deduce the equation. Use $A_r: \ce{C} = 12.0, \ce{Cu} = 63.5, \ce{O} = 16.0$.

First calculate the relative masses:

• $A_r(\ce{C}) = 12.0$
• $M_r(\ce{CuO}) = 63.5 + 16.0 = 79.5$
• $A_r(\ce{Cu}) = 63.5$
• $M_r(\ce{CO2}) = 12.0 + (2 \times 16.0) = 44.0$

Then convert every mass to moles:

Divide all values by the smallest, $0.100$:

$\ce{C} : \ce{CuO} : \ce{Cu} : \ce{CO2} = 1 : 2 : 2 : 1$

Therefore:

$\ce{C + 2CuO -> 2Cu + CO2}$

Check the symbols: there is one C atom, two Cu atoms and two O atoms on each side. Check the macroscopic data too: reactant mass is $1.20 + 15.9 = 17.1 \,\mathrm{g}$, and product mass is $12.7 + 4.40 = 17.1 \,\mathrm{g}$. Both checks support the equation for this complete data set.