When there is not sufficient of one reactant in a inter-base.netical reaction, the reaction stops abruptly. To number out the quantity of product produced, it have to be identified reactant will limit the inter-base.netical reaction (the limiting reagent) and also which reactant is in overfill (the overabundance reagent). One method of finding the limiting reagent is by calculating the amount of product that can be formed by each reactant; the one the produces less product is the limiting reagent.

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## Introduction

4 Tires + 2 Headlights = 1 Car

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figure 1: The synthetic reaction of make a car. Images used native Wikipedia through permission.

The initial problem is the there should be 4 tires come 2 headlights. The reactants have to thus occur in the ratio; otherwise, one will limit the reaction. There are 20 tires and also 14 headlights, therefore there space two methods of looking in ~ this problem. For 20 tires, 10 headlights are required, whereas because that 14 headlights, 28 tires space required. Due to the fact that there space not sufficient tires (20 tires is much less than the 28 required), tires space the limiting "reactant."

The limiting reagent is the reactant that is totally used up in a reaction, and thus determines once the reaction stops. Native the reaction stoichiometry, the precise amount that reactant essential to react with another element deserve to be calculated. If the reactants space not combined in the correct stoichiometric proportions (as suggested by the balanced inter-base.netical equation), then one of the reactants will certainly be completely consumed while one more will it is in left over. The limiting reagent is the one the is entirely consumed; it borders the reaction from continuing due to the fact that there is none left to react with the in-excess reactant.

There are two means to determine the limiting reagent. One an approach is to find and compare the mole ratio of the reactants offered in the reaction (approach 1). Another way is to calculation the grams of products produced native the given quantities of reactants; the reactant that produces the the smallest amount the product is the limiting reagent (approach 2).

How to uncover the Limiting Reagent: technique 2

Find the limiting reagent through calculating and comparing the lot of product every reactant will certainly produce.

Balance the inter-base.netical equation for the inter-base.netical reaction. Convert the provided information into moles. Usage stoichiometry because that each individual reactant to discover the massive of product produced. The reactant that produces a lesser quantity of product is the limiting reagent. The reactant the produces a larger amount the product is the excess reagent. To uncover the amount of remaining excess reactant, subtract the fixed of overfill reagent spend from the complete mass of overabundance reagent given.

b. If all of the 0.1388 moles of glucose were supplied up, over there would must be 0.1388 x 6 or 0.8328 mole of oxygen. Since there is an excess of oxygen, the glucose quantity is used to calculate the quantity of the products in the reaction.
Therefore, the mole proportion is: (0.8328 mol O2)/(0.208 mol C6H12O6)

This gives a 4.004 proportion of O2 come C6H12O6.

Step 4: Use the quantity of limiting reactant to calculation the amount of CO2 or H2O produced.

For carbon dioxide produced: $$\mathrm0.1388\: moles\: glucose \times \dfrac61 = 0.8328\: moles\: carbon\: dioxide$$.

Step 5: If necessary, calculate just how much is left in excess.

1.25 mol - 0.8328 mol = 0.4172 moles of oxygen left over

Step 5: The reactant the produces a bigger amount of product is the overabundance reagent
Step 6: Find the quantity of staying excess reactant by individually the massive of the overfill reagent spend from the full mass of overfill reagent given.

$$\mathrm2.40\:g\: Mg \times \dfrac1.00\: mol\: Mg24.31\:g\: Mg \times \dfrac1.00\: mol\: O_22.00\: mol\: Mg \times \dfrac32.0\:g\: O_21.00\: mol\: O_2 = 1.58\:g\: O_2$$ OR Mass of excess reagent calculated making use of the massive of the product: $$\mathrm3.98\:g\: MgO \times \dfrac1.00\: mol\: MgO40.31\:g\: MgO \times \dfrac1.00\: mol\: O_22.00\: mol\: MgO \times \dfrac32.0\:g\: O_21.00\: mol\: O_2 = 1.58\:g\: O_2$$ massive of full excess reagent provided – massive of excess reagent consumed in the reaction10.0g – 1.58g = 8.42g O2 is in excess.

Example $$\PageIndex3$$: Limiting Reagent

What is the limiting reagent if 76.4 grams the $$C_2H_3Br_3$$ to be reacted with 49.1 grams the $$O_2$$?

\<\ce4 C_2H_3Br_3 + 11 O_2 \rightarrow 8 CO_2 + 6 H_2O + 6 Br_2 \nonumber\>

Solution

Using technique 1:

A. $$\mathrm76.4\:g \times \dfrac1\: mole266.72\:g = 0.286\: moles\: of\: C_2H_3Br_3$$

$$\mathrm49.1\: g \times \dfrac1\: mole32\:g = 1.53\: moles\: of\: O_2$$

B. Presume that all of the oxygen is used up, $$\mathrm1.53 \times \dfrac411$$ or 0.556 mole of C2H3Br3 are required. Due to the fact that there are just 0.286 moles of C2H3Br3 available, C2H3Br3 is the limiting reagent.

Using approach 2:

$$\mathrm76.4\:g\: C_2H_3Br_3 \times \dfrac1\: mol\: C_2H_3Br_3266.72\:g\: C_2H_3Br_3 \times \dfrac8\: mol\: CO_24\: mol\: C_2H_3Br_3 \times \dfrac44.01\:g\: CO_21\: mol\: CO_2 = 25.2\:g\: CO_2$$

$$\mathrm49.1\:g\: O_2 \times \dfrac1\: mol\: O_232\:g\: O_2 \times \dfrac8\: mol\: CO_211\: mol\: O_2 \times \dfrac44.01\:g\: CO_21\: mol\: CO_2 = 49.1\:g\: CO_2$$

Therefore, by one of two people method, C2H3Br3is the limiting reagent.

A. $$\mathrm78\:g \times \dfrac1\: mol77.96\:g = 1.001\: moles\: of\: Na_2O_2$$

$$\mathrm29.4\:g \times \dfrac1\: mol18\:g= 1.633\: moles\: of\: H_2O$$

B. Assume that all of the water is consumed, $$\mathrm1.633 \times \dfrac22$$ or 1.633 mole of Na2O2 room required. Because there are just 1.001 mole of Na2O2, the is the limiting reactant.

Using method 2:

$$\mathrm78\:g\: Na_2O_2 \times \dfrac1\: mol\: Na_2O_277.96\:g\: Na_2O_2 \times \dfrac4\: mol\: NaOH2\: mol\: Na_2O_2 \times \dfrac40\:g\: NaOH1\: mol\: NaOH = 80.04\:g\: NaOH$$

Using either approach gives Na2O2 together the limiting reagent.

Example $$\PageIndex6$$: identifying the Limiting Reagent

Will 28.7 grams of $$SiO_2$$ react completely with 22.6 grams the $$H_2F_2$$? If not, determine the limiting reagent.

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Solution

A. $$\mathrm28.7\:g \times \dfrac1\: mole60.08\:g = 0.478\: moles\: of\: SiO_2$$

$$\mathrm22.6\:g \times \dfrac1\: mole39.8\:g = 0.568\: moles\: of\: H_2F_2$$

B. There have to be 1 mole of SiO2 because that every 2 moles of H2F2 consumed. Because the ratio is 0.478 to 0.568, 28.7 grams the SiO2 execute not react with the H2F2.

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C. Suspect that all of the silicon dioxide is offered up, $$\mathrm0.478 \times \dfrac21$$ or 0.956 moles of H2F2 space required. Due to the fact that there are just 0.568 moles of H2F2, that is the limiting reagent.