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explain stoichiometry and stoichiometric calculations with examples

Stoichiometry is a branch of chemistry that deals with the relative quantities of reactants and products in chemical reactions. Stoichiometry can be used to determine quantities such as the amount of products (in mass, moles, volume, etc.) that can be produced with given reactants and percent yield (the percentage of the given reactant that is made into the product).

Example
1.Fe + O2?Fe2O3
Therefore, we must balance the equation by placing coefficients before the various molecules and atoms to ensure that the number of atoms on the left side of the arrow corresponds exactly to the number of elements on the right.
4Fe +3O2?2Fe2O3.
The process of balancing an equation is basically trial and error. It gets easier and easier with practice. You will likely start to balance equations almost automatically in your mind.    

Almost all stoichiometric problems can be solved in just four simple steps:
1.Balance the equation.
2.Convert units of a given substance to moles.
3.Using the mole ratio, calculate the moles of substance yielded by the reaction.
4.Convert moles of wanted substance to desired units.

Problem 1: Given the following equation at STP:
N2(g) + H2(g)?NH3(g)    
Determine what volume of H2(g) is needed to produce 224 L of NH3(g).

Solution: 
Step 1: Balance the equation.

N2(g) + 3H2(g)?2NH3(g)    

Step 2: Convert the given quantity to moles. Note in this step, 22.4 L is on the denominator of the conversion factors since we want to convert from liters to moles. Remember your conversion factors must always be arranged so that the units cancel.
  224 L of NH3(g) * 1 mole / 22.4 L = 10 moles of NH3(g)    

Step 3: mole ratio.
  10 moles of NH3(g) * 3 moles H2(g) / 2 moles NH3(g) = 15 moles H2(g)    

Step 4, convert to desired units:
  15 moles H2(g) * 22.4 L / 1 mole = 336 L H2(g)    




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