Limiting and Excess Reactant

Limiting Reactant - the reactant in a chemical reaction that limits the amount of product that can be formed. The reaction will stop when all of the limiting reactant is consumed

Excess Reactant - the reactant in a chemical reaction that remains when a reaction stops when the limiting reactant is completely consumed. The excess reactant remains because there is nothing with which it can react

Limiting/Excess Reactant

No matter how many tires there are, if there are only 8 car bodies, then only 8 cars can be made

Likewise with chemistry - if there is only a certain amount of one reactant available for a reaction, the reaction must stop when that reactant is consumed whether or not the other reactant has been used up

The idea of excess

In a balanced equation like the one below, it is often assumed that all of the reacting chemicals change into products:

CaCO3 + 2HCl →CaCl2 + CO2 + H2O

In that case, at the end of the reaction, no CaCO3 or HCl will be left behind

However, if there is a shortage of, say, CaCO3 then the reaction will stop when the CaCO3 runs out

Some HCl will be left over, unable to react, as there is no more CaCO3

The HCl is said to be in excess

Summary

Chemical reaction equations give the ideal stoichiometric relationship among reactants and products

However, the reactants for a reaction in an experiment are not necessarily a stoichiometric mixture

In a chemical reaction, reactants that are not used up when the

reaction is finished are called excess reagents

Reagent that is completely used up is called the limiting reagent, because its quantity limit the amount of products formed

Why is this important?In lab or industry, we want to predict amount of product wecan expect from a reaction

Have to use stoichiometry : quantitative study of reactants and products in a reaction (mole to mole ratio)

When amounts of two reactants are given, we have to solve a limiting-reactant problem

Mole RatioThe mole ratio is the stoichiometric ratio of reactants and products and is the ratio of the coefficients for reactants and products found in the balanced chemical equation

in the reaction: 2 Mg(s) + O2(g) → 2MgO(s)

the mole ratio of: Mg : O2 : MgO

is: 2 : 1 : 2

That is, the complete reaction requires twice as many moles of magnesium as there are moles of oxygen

Example 1. A 2.00 g sample of ammonia is mixed with 4.00 g of oxygen. Which is the limiting reactant, how much product is produced and how much excess reactant remains after the reaction has stopped?

First, we need to create a balanced equation for the reaction:

4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)

2 g 4 g mx g

68 g 160 g 120 g

Next we can calculate how much product is produced by each reactant:

mx (by NH3) = (2 x 120) / 68 = 3.53 g

mx (by O2) = (4 x 120) / 160 = 3.00 g

The reactant that produces the lesser amount of product in this case is oxygen, which is thus the "limiting reactant"

Example 1.

4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)

2 g 4 g mx g

68 g 160 g 120 g

m(NH3) = (4 x 68) / 160 = 1.70 g NH3

Next, to find the amount of excess reactant, we must calculate how much of the non-limiting reactant (ammonia) actually did react with the limiting reactant (oxygen):

1.70 g is the amount of ammonia that reacted, not what is left over

To find the amount of excess reactant remaining, we have to subtract the amount that reacted from the amount in the original sample:

2.00 g (original sample) – 1.70 g (reacted) = 0.30 g NH3 remaining

x g

Stoichiometry of Precipitation Reactions

To calculate the volume of the given solution that would be required to completely precipitate the given ions from the solution

If you were to add salt (sodium chloride) to a solution of silver nitrate, the MOLECULAR reaction equation would be:

NaCl(aq) + AgNO3(aq) ↔ NaNO3(aq) + AgCl(s)

Use solubility Table to predict which compound will precipitate: AgCl is insoluble in water

we know that when NaCl and AgNO3 dissociate in water, they will

yield the following ions:

Na+(aq) + Cl−(aq) + Ag+(aq) + NO3−

(aq) 

Stoichiometry of Precipitation Reactions

MOLECULAR reaction equation:

NaCl(aq) + AgNO3(aq) ↔ NaNO3(aq) + AgCl(s)

While the NET ionic equation would be:

Na+(aq) + Cl-(aq) + Ag+(aq) + NO3-(aq) ↔ Na+(aq) + NO3

-(aq) + AgCl(s)

Then, write the OVERALL (complete) ionic equation:

Ag+(aq) + Cl-(aq) ↔ AgCl(s)since sodium and nitrate are spectator ions

to fully precipitate all of the silver(I), we need a 1:1 mole ratio of silver:chloride

Example 2. What volume of 0.05M Na3PO4 is required to precipitate all the silver

ions from 75ml of a 0.1M solution of AgNO3. Calculate the mass of

the precipitate formed

1. Determine the number of moles in the 0.1 Molar AgNO3 (acts as

limiting reactant):

Na3PO4 + 3AgNO3 → 3NaNO3 + Ag3PO4↓

x moles 0.0075 moles 0.0025 moles

1 mole 3 moles 1 mole

x = 0.0025 moles Na3PO4

n(AgNO3) = molarity x L of solution = 0.1 x 0.075 = 0.0075 moles

2. Write the balanced equation for the reaction:

3. Calculate the moles (or mmol) of the second reactant (Na3PO4)

Example 2.

1000 mL – 0.05 moles Na3PO4

x mL – 0.0025 moles Na3PO4

4. Now plug this molar amount of Na3PO4 into the molarity ratio using the

molarity of the solution given in the problem to determine the volume needed:

V = 50 mL 0.05M Na3PO4

5. Convert 0.0025 moles of precipitated silver phosphate into grams or other units, as required:

m = 0.0025 moles x 419 g/mole = 1.05 g Ag3PO4

Na3PO4 + 3AgNO3 → 3NaNO3 + Ag3PO4

x moles 0.0075 moles 0.0025 moles

1 mole 3 moles 1 mole

Kiek ml 8% H2SO4 tirpalo, kurio tankis = 1,055 g/cm3 reikės, norint nusodinti visus bario jonus, esančius 10 g bario chlorido?

Example 3.

BaCl2 + H2SO4 BaSO4 + 2HCl

10 g x

208 g 98 g 2BaClM 137 + 235,5 = 208 g/mol

42SOHM 21 + 32 + 416 = 98 g/mol

Apskaičiuojame bario chlorido molių skaičių:

208

102 M

mnBaCl 0,048 mol

Reakcijoje dalyvaus tiek pat molių H2SO4. Apskaičiuojame, kiek gramų grynos sieros rūgšties dalyvaus reakcijoje:

98048,0Mnm4242 SOHSOH 4,70 g

Apskaičiuojame, kiek reikės gramų 8% sieros rūgšties tirpalo:

8 g H2SO4 – 100 g tirpalo

4,7 g H2SO4 – x x = 8

1007,458,75 g

Reikiamą 8% sieros rūgšties tirpalo tūrį apskaičiuojame tirpalo masę padalindami iš tankio:

055,1

75,58mV t 55,68 ml 55,7 ml

Atsakymas. Norint nusodinti visus bario jonus, esančius 10 g BaCl2, reikės paimti 55,7 ml 8% H2SO4 tirpalo

Example 3.