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How would you identify a limiting reagent in a chemical reaction?

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Last updated date: 24th Feb 2024
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IVSAT 2024
Answer
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Hint:In a chemical reaction, there are two types of reagents present.one is excess reagent which is not consumed completely in the reaction and another one is consumed first and the product is limited by that reagent.
Limiting reagent: The reactant in a chemical reaction which is completely consumed when the reaction is completed. How much product will be formed is determined by the limiting reagent.

Complete step-by-step answer:We can find out the limiting reagent by a trick or you can say a formula that is whose the ratio of mole to stoichiometric coefficient is low is known as limiting reagent.
Some rules should be followed to find out the limiting reagent:
(1) Balance the equation.
(2) Convert it into the form of mole.
(3) Calculate the ratio of mole to stoichiometric coefficient.
(4) Calculate the mass of the product.
(5) You can calculate the remaining excess reagent also.
To understand this let us take an example, $1g$hydrogen gas reacts with$1g$chlorine gas and we have to find out the limiting reagent and weight of the product.
${H_2}(g) + C{l_2}(g) \to 2HCl$
In our example, the equation is already balanced.
Moles of hydrogen${n_h} = \dfrac{w}{{{M_h}}}$putting the values we get,
$ = \dfrac{1}{2} = 0.5mole$
Stoichiometric coefficient of hydrogen is$1$. Therefore the ratio of moles to S.C. will be:
${(\dfrac{n}{{S.C.}})_{{H_2}}} = \dfrac{{1/2}}{1} = \dfrac{1}{2}$
For chlorine gas, the number of moles $ = \dfrac{1}{{71}}moles$and S.C. for chlorine is$1$. Therefore the ratio of moles to S.C. will be:
${(\dfrac{n}{{S,C,}})_{C{l_2}}} = \dfrac{{1/71}}{1} = \dfrac{1}{{71}}$
The ratio of chlorine is less than the hydrogen. So, the limiting reagent is chlorine. If one mole of chlorine gives us two mole of $HCl$then, $\dfrac{1}{{71}}mole$of chlorine gives us $\dfrac{2}{{71}}mole$of$HCl$.therefore the moles of $HCl$is$\dfrac{2}{{71}}mole$and molecular mass of$HCl$is$36.5u$. We know that $n = \dfrac{m}{M}$Where $n = $no. of moles
$m = $Mass
$M = $Molecular mass, putting all the values in this expression. We get,
$
  \dfrac{2}{{17}} = \dfrac{m}{{36.5}} \\
   \Rightarrow m = \dfrac{2}{{17}} \times 36.5 \\
   = \dfrac{{73}}{{71}}g = 1.02g \\
 $

Note:While finding the limiting reagent, be careful with the number of moles. The formula of no. of moles and molar mass should be remembered. Always start numerical with balancing the chemical equation. If you will not balance the equation the value of mass of the product will be wrong.
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