Question

How can we do Stoichiometry chemistry of the mole problem?

Answer 1

In order to solve the stoichiometry of a chemical reaction we have to calculate the number of moles of reactants and products in the chemical reaction. As it is simply the calculation of reactants and products in the chemical reaction.

Complete step by step answer:
Stoichiometry gives quantitative relations. This is also used to determine the amount of product and reactions that are needed in a reaction.
In Gas Stoichiometry reactions involve gases, gases are dealing in known pressures, temperature, volume.
Stoichiometry is not only used for balancing the chemical equation but also needed in conversions. That is converting from moles using moles mass as the conversion factor, or form grows to milliliters using density.
Converting grams to mules: Stoichiometry is not only used to balance the chemical equation here we are taking an example of conversion of grams into moles.
Suppose I find an amount of $\text{NaCl}$ (Sodium Chloride) in $2.00$ grams.
$\dfrac{2.00\,\text{gram NaCl}}{58.44\,\text{gram NaCl mo}{{\text{l}}^{-1}}}$ will after saving $=0.034$ moles.
Hence in grams of Sodium Chloride there is $0.034$ mole present.
Examples: The Stoichiometry the after used to balance chemical equations for examples
${{\text{N}}_{2}}+3{{\text{H}}_{2}}\to 2\text{N}{{\text{H}}_{3}}$
The Stoichiometry of the reaction tells that one mole of nitrogen gas reacting with three moles of hydrogen gives two moles of ammonium gas. The molar rather for the reaction will.
$1:3:2$
Means $1$ mole of Nitrogen gas: Three moles of Hydrogen gas: two moles of Ammonium gas. Simple in Stoichiometry of chemical reaction we have to convert as quantity into molar form
In the chemical industry a knowledge of Stoichiometry is essential for the calculation of the yields of chemical products.
Suppose we are taking an example that sulphur dioxide reacts with: with oxygen gas the form sulphur trioxide.
If $0.4$ moles of sulphur Dioxide reacts with excess oxygen gas, how many moles of Sulphur triored will form
For this first we have to which chemical reaction
$2\text{S}{{\text{o}}_{2}}+{{\text{O}}_{2}}\to \text{2S}{{\text{o}}_{3}}$
For balancing the equation the molar reaction will $2:1 :2$
Means the $2$ moles of $\text{S}{{\text{o}}_{2}}$ sulfur dioxide give the same two moles of product (Sulphur trioxide) on solution.
$\dfrac{6.4}{1}\times \dfrac{1\,\text{molor of }{{\text{O}}_{2}}}{2}=\dfrac{6.4}{2}\text{mole of }{{\text{O}}_{2}}$
$=3.2$ moles of oxygen required
In the above reaction when the $6.4$moles above Sulphur dioxide react with $3.2$ moles of oxygen given the $6.4$ moles of Sulfur trioxide.

Note: The term stoichiometry is derived from the ancient greek word which means element and measure. It depends on the basic laws such as law of conservation of mass, law of definite proportions and so on.