Transcript ATLIEKŲ TVARKYMAS

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 we
can 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 Ratio
The 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:
the mole ratio of:
is:
2 Mg(s) + O2(g) → 2MgO(s)
Mg : O2 :
2
: 1
:
MgO
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:
2g
4g
mx g
4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(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.
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):
2g xg
4g
mx g
4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)
68 g
160 g
120 g
m(NH3) = (4 x 68) / 160 = 1.70 g NH3
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
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)
Then, write the OVERALL (complete) ionic equation:
Na+(aq) + Cl-(aq) + Ag+(aq) + NO3-(aq) ↔ Na+(aq) + NO3-(aq) + AgCl(s)
While the NET ionic equation would be:
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):
n(AgNO3) = molarity x L of solution = 0.1 x 0.075 = 0.0075 moles
2. Write the balanced equation for the reaction:
x moles
0.0075 moles
0.0025 moles
Na3PO4 + 3AgNO3 → 3NaNO3 + Ag3PO4↓
1 mole
3 moles
1 mole
3. Calculate the moles (or mmol) of the second reactant (Na3PO4)
x = 0.0025 moles Na3PO4
Example 2.
x moles
0.0075 moles
0.0025 moles
Na3PO4 + 3AgNO3 → 3NaNO3 + Ag3PO4
1 mole
3 moles
1 mole
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:
1000 mL – 0.05 moles Na3PO4
x mL – 0.0025 moles Na3PO4
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
Example 3.
Kiek ml 8% H2SO4 tirpalo, kurio tankis  = 1,055 g/cm3 reikės, norint
nusodinti visus bario jonus, esančius 10 g bario chlorido?
10 g
x
BaCl2 + H2SO4  BaSO4 + 2HCl
208 g
98 g
M BaCl2  137 + 235,5 = 208 g/mol
M H2SO4  21 + 32 + 416 = 98 g/mol
Apskaičiuojame bario chlorido molių skaičių:
n BaCl2
m
10


 0,048 mol
M 208
Reakcijoje dalyvaus tiek pat molių H2SO4. Apskaičiuojame, kiek
gramų grynos sieros rūgšties dalyvaus reakcijoje:
mH2SO4  n  MH2SO4  0,048 98  4,70 g
Example 3.
Apskaičiuojame, kiek reikės gramų 8% sieros rūgšties tirpalo:
8 g H2SO4
– 100 g tirpalo
4,7 g H2SO4 – x
4,7  100
 58,75 g
x=
8
Reikiamą 8% sieros rūgšties tirpalo tūrį apskaičiuojame tirpalo
masę padalindami iš tankio:
m t 58,75
V

 55,68 ml  55,7 ml

1,055
Atsakymas. Norint nusodinti visus bario jonus, esančius 10 g BaCl2,
reikės paimti 55,7 ml 8% H2SO4 tirpalo