The limiting reagent It is the one that is completely consumed and determines how much mass of products are formed in a chemical reaction; while the excess reagent is one that does not react completely after the limiting reagent has been consumed.
In many reactions, an excess of a reagent is looked for to ensure that all of the reagent of interest reacts. For example, if A reacts with B to produce C, and it is desired that A react completely, an excess of B is added. However, synthesis, and scientific and economic criteria, are what decide whether an excess of A is appropriate. or from B.
The limiting reagent determines the amount of product that can be formed in the chemical reaction. Therefore, if it is known how much of A reacted, it is immediately determined how much of C was formed. The excess reactant never reveals the amounts of product formed..
What if both A and B are consumed in the reaction? Then we speak of an equimolar mixture of A and B. In practice, however, it is not an easy task to ensure that there are equal numbers of moles or equivalents of all reactants; In this case, either of the two, A or B, can be used to calculate the amount of C formed.
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There are many ways to identify and calculate the amount of the limiting reagent that can be involved in the reaction. Once calculated, the other reactants are in excess.
A method that allows identifying which is the limiting reagent, based on the comparison of the proportion of reagents with the stoichiometric ratio, is the one described below.
A chemical reaction can be outlined as follows:
aX + bY => cZ
Where X, Y and Z represent the number of moles of each reactant and product. While, a, b and c represent their stoichiometric coefficients, resulting from the chemical balance of the reactions.
If the quotient (X / a) and the quotient (Y / b) are obtained, the reactant with the lower quotient is the limiting reactant.
When the indicated ratios are calculated, the relationship between the number of moles present in the reaction (X, Y and Z) and the number of moles involved in the reaction is being established, represented by the stoichiometric coefficients of the reactants (a and b).
Therefore, the lower the quotient indicated for a reagent, the greater the deficit of that reagent to complete the reaction; and therefore, it is the limiting reactant.
Yestwo(s) + 3 C (s) => SiC (s) + 2 COtwo(g)
3 g of SiO are reactedtwo (silicon oxide) with 4.5 g of C (carbon).
Moles of SiOtwo
Mass = 3 g
Molecular weight = 60 g / mol
Number of moles of SiOtwo = 3g / (60g / mol)
0.05 moles
Number of moles of C
Mass = 4.5 g
Atomic weight = 12 g / mol
Number of moles of C = 4.5 g / (12g / mol)
0.375 moles
Quotient between the number of moles of the reactants and their stoichiometric coefficients:
For SiOtwo = 0.05 moles / 1 mole
Quotient = 0.05
For C = 0.375 moles / 3 moles
Quotient = 0.125
From the comparison of the values of the quotients, it can be concluded that the limiting reactant is SiOtwo.
The mass produced of SiC is calculated from the previous reaction, when 3 g of SiO is usedtwo and when using the 4.5 g of C
(3 g SiOtwo) x (1 mol SiOtwo/ 60 g SiOtwo) x (1 mol SiC / 1 mol SiOtwo) X (40 g SiC / 1 mol SiC) = 2 g SiC
(4.5 g C) x (3 mol C / 36 g C) x (1 mol SiC / 3 mol C) x (40 g SiC / 1 mol SiC) = 5 g SiC
So, more SiC (silicon carbide) would be produced if the reaction occurred by consuming all the carbon than the amount produced by consuming all of the SiOtwo. In conclusion, the SiOtwo is the limiting reagent, since when all the excess C is consumed, more SiC would be generated.
0.5 moles of aluminum are reacted with 0.9 moles of Chlorine (Cltwo) to form aluminum chloride (AlCl3): What is the limiting reactant and what is the excess reactant? Calculate the mass of the limiting reagent and the excess reagent
2 Al (s) + 3 Cltwo(g) => 2 AlCl3(s)
The quotients between the moles of the reactants and the stoichiometric coefficients are:
For aluminum = 0.5 moles / 2 moles
Aluminum quotient = 0.25
For Cltwo = 0.9 moles / 3 moles
Cl quotienttwo = 0.3
Then the limiting reagent is aluminum.
A similar conclusion is reached if the moles of chlorine required to combine with the 0.5 moles of aluminum are determined.
Moles of Cltwo = (0.5 moles of Al) x (3 moles of Cltwo/ 2 moles of Al)
0.75 moles of Cltwo
Then there is an excess of Cltwo: 0.75 moles are required to react with aluminum, and 0.9 moles are present. Therefore, there is an excess of 0.15 moles of Cltwo.
It can be concluded that the limiting reagent is aluminum
Limiting reagent mass:
Mass of aluminum = 0.5 moles of Al x 27 g / mole
13.5 g.
The atomic mass of Al is 27g / mol.
Mass of excess reagent:
0.15 moles of Cl remainedtwo
Cl masstwo excess = 0.15 moles of Cltwo x 70 g / mol
10.5 g
The following equation represents the reaction between silver nitrate and barium chloride in aqueous solution:
2 AgNO3 (ac) + BaCltwo (ac) => 2 AgCl (s) + Ba (NO3)two (ac)
According to this equation, if a solution containing 62.4g of AgNO3 is mixed with a solution containing 53.1 g of BaCltwo: a) What is the limiting reagent? b) How many of which reactant remain unreacted? c) How many grams of AgCl were formed?
Molecular weights:
-AgNO3: 169.9g / mol
-BaCltwo: 208.9 g / mol
-AgCl: 143.4 g / mol
-Bath3)two: 261.9 g / mol
To apply Method 1, which allows the identification of the limiting reagent, it is necessary to determine the moles of AgNO3 and BaCltwo present in the reaction.
AgNO moles3
Molecular weight 169.9 g / mol
Mass = 62.4 g
Number of moles = 62.4 g / (169.9 g / mol)
0.367 moles
Moles of BaCltwo
Molecular weight = 208.9 g / mol
Mass = 53.1 g
Number of moles = 53.1 g / (208.9 g / mol)
0.254 moles
Determination of the quotients between the number of moles of the reactants and their stoichiometric coefficients.
For AgNO3 = 0.367 moles / 2 moles
Quotient = 0.184
For the BaCltwo = 0.254 moles / 1 mole
Quotient = 0.254
Based on Method 1, the value of the ratios allows to identify AgNO3 as the limiting reagent.
The stoichiometric balance of the reaction indicates that 2 moles of AgNO3 react with 1 mole of BaCltwo.
Moles of BaCltwo= (0.367 moles of AgNO3) x (1 mol BaCltwo/ 2 moles of AgNO3)
0.1835 moles of BaCltwo
And the moles of BaCltwo that did not intervene in the reaction, that is, that are in excess are:
0.254 moles - 0.1835 moles = 0.0705 moles
BaCl masstwo in excess:
0.0705 mol x 208.9 g / mol = 14.72 g
Resume:
Excess reagent: BaCltwo
Excess mass: 14.72 g
To calculate the mass of the products, the calculations are made based on the limiting reagent.
g AgCl = (62.4 g AgNO3) x (1 mol AgNO3/ 169.9 g) x (2 mol AgCl / 2 mol AgNO3) x (142.9 g / mol AgCl)
52.48 g
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