Question

tell me about limiting reagent with solve the example

Answer 1

Before solving a problem let me share an algebraic approach of solving problems regarding stoichiometry.
$\text{Notation}:\\n=\text{Number of moles}\\m=\text{Mass of substance}\\M=\text{Molar mass of substance}\\N=\text{Number of particles}\\N_A=\text{Avogadro's Number}\\V=\text{Volume}\\\bar{V}=\text{Molar Volume of gas}\\\begin{cases}=22.4 \text{ L/mol at old S.T.P}\\=22.7 \text{ L/mol at new S.T.P}\\=24 \text{ L/mol at N.T.P}\end{cases}\\v=\text{Stoichiometric Coefficient}$
Now, use the following formulae for calculating one quantity, given other quantities. This formula works as each term in the numerator is proportional to other term in numerator.
$n=\frac{m}{M}=\frac{N}{N_A}=\frac{V}{\bar{V}}\qquad :eq(1)$
In any chemical reaction the following `rule' always holds true: 
$\frac{n_i}{v_i}=Constant$
where $n_i$ and $v_i$ are the number of moles and stoichiometric coefficient of $i^{th}$ species in a reaction.(which is not an excess reagent).
As an example consider the simple chemical reaction as follows:$aA+bB(excess)\longrightarrow cC+dD\\then,\\\frac{n_A}{a}=\frac{n_C}{c}=\frac{n_D}{d}$
You can pick any two chemical species (say, A and C) and equate the $\frac{n_i}{v_i}$ ratio, express $n_i$ in terms of any of the quantities in $eq(1)$, solve for the unknown quantity and your question is solved. 
Now, let us understand how to identify the limiting reagent with certainty. First of all using $eq(1)$, convert all information about the reactant species in terms of  moles. You just have to compare the $\frac{n_i}{v_i}$ ratio of all the reactants and the species having the least value of this ratio, will be the limiting reagent. Keep in mind that you cannot equate this ratio of any excess reagents to any other species. Let us solve an example using this strategy,
$Q)$Consider respiration, one of the most common chemical reactions on earth.
$C_6H_{12}O_6+6O_2\longrightarrow 6CO_2+6H_2O+energy$
What mass of of carbon dioxide forms in the reaction of $25g$ of glucose with $40g$ of oxygen?
$\dfrac{n_{C_6H_{12}O_6}}{v_{C_6H_{12}O_6}}=\dfrac{m_{C_6H_{12}O_6}}{1\cdot M_{C_6H_{12}O_6}}=\frac{25}{180}=\frac{5}{36}\\\dfrac{n_{O_2}}{v_{O_2}}=\dfrac{m_{O_2}}{6\cdot M_{O_2}}=\frac{40}{6\cdot32}=\frac{5}{24}\\\text{As you can see,}\\\dfrac{n_{C_6H_{12}O_6}}{v_{C_6H_{12}O_6}}<\dfrac{n_{O_2}}{v_{O_2}}$
Therefore, Glucose is the limiting reagent. Now, equate this ratio for Glucose and Carbon Dioxide.
$\dfrac{n_{C_6H_{12}O_6}}{v_{C_6H_{12}O_6}}=\dfrac{n_{CO_2}}{v_{CO_2}}\\\Rightarrow \dfrac{5}{36}=\dfrac{m_{CO_2}}{6\cdot M_{CO_2}}\\\Rightarrow \dfrac{5}{36}=\dfrac{m_{CO_2}}{6\cdot44}\\\Rightarrow m_{CO_2}=36.67g$