The quantitative study of the relative amounts of reactants and products in chemical reactions is referred to as **stoichiometry**. Stoichiometric problems are solved in just four simple steps which are as following:

- First step is the balance the equation
- Convert units of a given reactants and products into moles.
- This is followed by calculating the moles of substance yielded by the reaction using the mole ratio,
- The final step is to convert moles of desired substance to desired units.

$Fe + O_2 \rightarrow Fe_{2}O_{3}$

As mentioned, the first step in stoichiometric calculation is the balancing the equation. As we know that reactants of a chemical equation are never destroyed or lost and therefore, the products of a reaction exactly correspond to the original reactants. Now, let’s balance the given equation:

$Fe + O_2 \rightarrow Fe_{2}O_{3}$

According to the equation, 1 iron (Fe) atom react with two oxygen (O) atoms to obtain 2 iron atoms and 3 oxygen atoms; the reaction is not balanced.

$4Fe + 3O_{2} \rightarrow 2Fe_{2}O_{3}$

According to the above mentioned balanced equation, 4 atoms of iron and 6 atoms of oxygen (since 3 $\times$ 2 = 6 ) react to form 4 iron (since 2 $\times$ 2 = 4 ) and 6 oxygen (2 $\times$ 3 = 6). The atoms on both sides of the equation match. Now, convert the given units to moles using conversion factors and then calculate the moles of substance.

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- One mole (1 mol) of an element = 6.022 $\times$ 10$^{23}$ entities (the Avogadro Number).

Therefore, 1 mol of carbon-12 means there are 6.022 $\times$ 10$^{23}$ carbon-12 atoms; 1 mol of H$_{2}$O contains 6.022 $\times$ 10$^{23}$ H$_{2}$O molecules.

- Molar mass of an element = mass(g)/mole.

- Molar mass of an compound = Sum total of molar masses of atoms of elements in the formula

- Mole to mass stoichiometry formula: Mass (g) = $\frac{no. \ of \ moles \times no. \ of \ grams}{1 \ mol}$

- Mass to mole stoichiometry formula: No. of moles = $\frac{Mass (g) \times \ 1 \ mol}{no \ of \ grams}$

- Moles to substance entity (atoms, molecules etc) formula: No. of entities = $\frac{no. \ of \ moles \times \ 6.022 \times 10^{23} \ entities}{1 \ mol}$

- Substance entity to mole stoichiometry formula: No. of moles = $\frac{no. \ of \ entities \times 1 \ mol}{6.022 \times 10^{23} \ entities}$

- Mass percent of an element in a compound is calculated using molecular mass and chemical formula.

The formula is as follows:Mass % of element X = $\frac{atoms \ of \ X \ in \ formula \times atomic \ mass \ of \ X \ (amu)}{molecular (or formula) \ mass \ of \ compound \ (amu)}$ $\times$ 100Since the chemical formula provides information about moles of each element in a compound, the mass percent of element X can be determined using the molar mass. The stoichiometry formula is as followsMass % of element X = $\frac{moles \ of \ X \ in \ formula \times molar \ mass \ of \ X \ (amu)}{mass \ of \ one \ mole \ of \ compound (amu)}$ $\times$ 100.

The mole to mole

Now, let’s suppose, we want to calculate the number of mole produced by 2 moles of oxygen. For the purpose, the following equation will be used

$H_{2} + O_{2} \rightarrow H_{2} O$.

Now, let’s first balance the equation. By balancing the equation, we get following one.

$2H_{2} + O_{2} \rightarrow H_{2} O$.

Here, we can see one mole of oxygen produces two moles of water. The stoichiometric ratio is 1 mole of oxygen: 2 mole of water. Let’s assume that there is abundant hydrogen gas available for reaction. Under this condition,

$\frac{2 \ mole \ of \ oxygen \times 2 \ mole \ of \ water}{1 \ mole \ of \ oxygen}$ = 4 moles of water

It uses

Example: For example, convert 20 grams of water to moles of water. The molar mass of water is 20 g/mol. Therefore the molecular weight of water in g/mol can be used to convert gram to mole. Thus,

$\frac{20 \ g \ water \times 1 \ mol}{20 g \ water}$ = 1 mol water.

According to the “ideal gas law”, at

The equation for gas law is given as following:

P V = n R T.

Here

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