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# limiting reagent calculator

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Limiting reagent

In a chemical reaction, the limiting reagent, also known as the "limiting reactant", is the substance which is totally consumed when the chemical reaction is complete. The amount of product formed is limited by this reagent since the reaction cannot proceed further without it. The other reagents may be present in excess of the quantities required to react with the limiting reagent.

The limiting reagent must be identified in order to calculate the percentage yield of a reaction, since the theoretical yield is defined as the amount of product obtained when the limiting reagent reacts completely.

Given the balanced chemical equation which describes the reaction, there are several equivalent ways to identify the limiting reagent and evaluate the excess quantities of other reagents.

## Method 1: Comparison of reactant amounts

This method is most useful when there are only two reactants. One reactant (A) is chosen, and the balanced chemical equation is used to determine the amount of the other reactant (B) necessary to react with A. If the amount of B actually present exceeds the amount required, then B is in excess and A is the limiting reagent. If the amount of B present is less than required, then B is the limiting reagent.

### Example for two reactants

Consider the combustion of benzene, represented by the following chemical equation:

2 C_6H_6 + 15 O_2 \rightarrow 12 CO_2 + 6 H_2O

This means that 15 mol molecular oxygen (O_2) is required to react with 2 mol benzene (C_6H_6).

The amount of oxygen required for other quantities of benzene can be calculated using cross-multiplication (the rule of three). For example, if 1.5 mol C_6H_6 is present, 11.25 mol O_2 is required:

1.5 \ \mbox{mol}\,C_6H_6 \times \frac{15 \ \mbox{mol}\,O_2}{2 \ \mbox{mol}\,C_6H_6} = 11.25 \ \mbox{mol}\,O_2\

If in fact 18 mol O_2 are present, there will be an excess of (18 - 11.25) = 5.75 mol of unreacted oxygen when all the benzene is consumed. Benzene is then the limiting reagent.

This conclusion can be verified by comparing the mole ratio of O_2 and C_6H_6 required by the balanced equation with the mole ratio actually present:

required: \frac{mol O_2}{mol C_6H_6} = \frac{15 mol O_2}{2 mol C_6H_6}=7.5 mol O_2
actual: \frac{mol O_2}{mol C_6H_6} = \frac{18 mol O_2}{1.5 mol C_6H_6}=12 mol O_2

Since the actual ratio is larger than required, O_2 is the reagent in excess, which confirms that benzene is the limiting reagent.

## Method 2: Comparison of product amounts which can be formed from each reactant

In this method the chemical equation is used to calculate the amount of one product which can be formed from each reactant in the amount present. This method can be extended to any number of reactants more easily than the first method.

### Example

Which reactant is limiting if 20.0 g of iron (III) oxide (Fe2O3) are reacted with 8.00 g aluminium (Al) in the followng thermite reaction?

Fe_2O_3(s) + 2 Al(s) \rightarrow 2 Fe(l) + Al_2O_3(s)\,

Since the reactant amounts are given in grams, they must be first converted into moles for comparison with the chemical equation, in order to determine how many moles of Fe can be produced from either reactant.

Moles produced of Fe from reactant Fe_2O_3

mol Fe_2O_3 = \frac{grams Fe_2O_3}{g/mol Fe_2O_3}\,
mol Fe_2O_3 = \frac{20.0 g}{159.7 g/mol} = 0.125 mol\,
mol Fe = 0.125 \ \mbox{mol}\,Fe_2O_3 \times \frac{2 \ \mbox{mol}\,Fe}{1 \ \mbox{mol}\,Fe_2O_3} = 0.250\ \mbox{mol}\,Fe\

Moles produced of Fe from reactant Al

mol Al = \frac{grams Al}{g/mol Al}\,
mol Al = \frac{8.00 g}{26.98 g/mol} = 0.297 mol\,
mol Fe = 0.297 \ \mbox{mol}\,Al \times \frac{2 \ \mbox{mol}\,Fe}{2 \ \mbox{mol}\,Al} = 0.297\ \mbox{mol}\,Fe\

There is enough Al to produce 0.297 mol Fe, but only enough Fe2O3 to produce 0.250 mol Fe. This means that the amount of Fe actually produced is limited by the Fe2O3 present, which is therefore the limiting reagent.

### Shortcut

It can be seen from the example above that the amount of product (Fe) formed from each reagent X (Fe2O3 or Al) is proportional to the quantity

\frac{\mbox{Moles of Reagent X }}{\mbox{Coefficient of Reagent X}}

This suggests a shortcut which works for any number of reagents. Just calculate this formula for each reagent, and the reagent that has the lowest answer to this formula is the limiting reagent.

Question:Adipic acid is treated with excess thionyl chloride, and the resulting acid dichoride is reacted with 20mL butyl amine (d=0.74). After work up and recrystallization, 12.5 g of N,N'-dibutyl adipamide is obtained.

Answers:You can calculate the theoretical yield of the product based on the original mass of adipic acid. You'll need to look at the equations, but I'm guessing there is a 2:1 relationship between moles of adipic acid and your amide product. So, grams adipic acid --> moles adipic acid --> moles product --> grams product. Then, calculate moles of butyl amine. Volume X density = grams --> moles dibutyl amine --> moles product --> grams product. Between those two, the one which gives you the smaller mass of product is your limiting reactant, and the mass of product would be your theoretical yield. Dividing 12.5 grams by the theoretical yield, and multiplying by 100 will give you the % yield. Hope this helps...

Question:In order to calculate the heat of this reaction, I need to know the moles of limiting reagent I used, but I don't know which is the limiting reagent: CH3-COOH or NH3??? 30 mL of each were used

Answers:depends on how much you start with of each material

Question:If 25 grams of sodium bromide (NaBr) are mixed with an excess amount of potassium chloride (KCl), what will the theoretical yield of sodium chloride be? The molar mass of NaBr is 102.9 g/1 mol and the molar mass of KCl is 75.45 g/1 mol. I know the formula for percent yield is actual yield / theoretical yield x 100% but I don't know how to get to that step. Please any help would be much appreciated!

Answers:25 g NaBr x 1 mol NaBr/ 102.9 g NaBr x 1 mol NaCl/ 1 mol NaBr = 0.2430 moles of NaCl this would be the theoretical yield in moles. if they want the answer in grams then you'd have to switch it back to grams. 0.2430 moles NaCl x 75.45 g NaCl/ 1 mol NaCl = 18.33 g NaCl

Question:Hydrogen peroxide is used as a cleaning agent in the treatment of cuts and abrasions for several reasons. It is an oxidizing agent that can directly kill many microorganisms; it decomposes upon contact with blood, releasing elemental oxygen gas (which inhibits the growth of anaerobic microorganisms); and it foams upon contact with blood, which provides a cleansing action. In the laboratory, small quantities of hydrogen peroxide can be prepared by the action of an acid on an alkaline earth metal peroxide, such as barium peroxide. BaO2(s) + 2HCl(aq)H2O2(aq) + BaCl2(aq) What amount of hydrogen peroxide should result when 1.45 g of barium peroxide is treated with 25.8 mL of hydrochloric acid solution containing 0.0272 g of HCl per mL? g H2O2 I got 12.0 g H2O2 is this correct??

Answers:BaO2(s) + 2HCl(aq) -------> H2O2(aq) + BaCl2(aq) Molar mass of BaO2 = 169.3 g/mol Molar mass of HCl = 36.45 g/mol Mass of HCl in solution: (0.0272 g/mL)(25.8 mL) = 0.70 g HCl Mole BaO2 = 1.45 g / 169.3 g/mol = 0.0086 mol Mole HCl = 0.70 g / 36.45 mol = 0.019 mol According to the equation, 1 mol BaO2 reacts with 2 mol HCl 0.0086 mol BaO2 reacts with 2x 0.0086 = 0.0172 mol HCl. Since we have more than thisamount (0.019 mol), HCl is excess and BaO2 is limiting. The amount of H2O2 depends on BaO2. Since 1 mol BaO2 produces 1 mol H2O2 0.0086 mol BaO2 will produce 0.0086 mol H2O2. Molar mass of H2O2 = 34 g /mol Mass of H2O2 = 0.0086 mol x 34 g/mol = 0.2924 g