A balanced chemical reaction equation provides the reaction ratios for all the reacting materials involved in the given chemical change. The* laws of conservation of mass *which determines the amounts of reactants or products which are present in the chemical change. The **reaction stoichiometry** looks into the aspects of basics of chemical change, the limiting reactant concept, the theoretical yield, the actual yield and the percent yield.

The reaction ratios are the key aspects of* reaction stoichiometry* and till the reaction is balanced we can make a reaction ratio for any reactant to reactant, reactant to product and even product to product essential for the stoichiometry.

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Any chemical process can be described by a single stoichiometry equation which defines the molar ratio of the reacting species and reaction products. This is also known as overall reaction equation. In real chemical systems corresponding to such a single chemical reaction or rather when a reactant react with each other forming products immediately are not very common and in most of the cases the reaction of the reactants produce intermediates and these intermediates react with each other and finally the products are seen after many coupled reaction steps.

Each individual steps is called* elementary reaction* and within each elementary reactions there are no macroscopically observable intermediate between reactants and products. For any chemical reactions the starting amounts are finite and when one of the reactants is completely consumed by the reaction, the reaction cannot proceed any further and hence the determining products amount can be calculated. In other words the amount of reactants limits the amount of products formed.

The amount of products calculated using limiting reactant concept is considered as theoretical yield. The amount of products which are experimentally formed in a reaction is considered to be the actual yield. There is a difference between actual and theoretical yield as the chemical reactions do not always produce the maximum possible amount of products.

**Percent yield or % yield = $\left[\frac{(actual yield)}{(theoretical yield)}\right]$ $\times$ 100 %. **

The percent yield formula rule is used to determine the theoretical yield of a particular reaction.

There are many chemical reaction processes where a single overall reaction is never found to suffix the chemical change. The stoichiometry of oxidation of hydrocarbons sourced from exhaust gases in the troposphere cannot be described by a single reaction and many of the hydrocarbons emitted to troposphere and the change in ratio dependent on the type of pollution source.

Qualitatively the rate of reaction is easy way to understand as it represent the change in concentration of the reactants as well as products. When we deal with rates of reaction quantitatively we need specific* reaction stoichiometry. *The rate of reaction can be expressed as the change in concentration of either reactants or products as a function of time.

The numerical values of the rate of formation of reactants and products are related but different due to reaction stoichiometry. The reaction for rate is given out by dividing each rate by its stoichiometry coefficient.

If we have a reaction of this nature A is decomposing into two compounds then we have a working relation of the reaction as:

**Rate of appearance for C will be given out as **

Rate = $\frac{\Delta [C_{2}]}{\Delta t}$ = $\frac{1}{2}$ $\frac{[\Delta A_{2}O_{5}]}{\Delta t}$ = $\frac{1}{4}$ $\frac{[BO_{2}]}{\Delta t}$

There are different information in the equation mentioned above. It gives the relation between the rates of change of concentration of reactants and products.

Based on** reaction stoichiometry **the rate of appearance of $C_2$ is one half the rate of disappearance of $A_{2}O_{5}$ and then one fourth of $BO_{2}$ appearance.

Rate = $\frac{1}{2}$ $\frac{[\Delta A_{2}O_{5}]}{\Delta t}$ = k $A_{2}O_{5}$

Or, rate = $\frac{[\Delta A_{2}O_{5}]}{\Delta t}$ = 2k $A_{2}O_{5}$

The integrated equation should look like: ln $\frac{[A_{2}O_{5}]t}{[A_{2}O_{5}]0}$ = -2kt

Hence the half-life equation would be $t_{1/2}$ = $\frac{0.693}{2k}$

**Differential rate equation:**

- $\frac{\Delta [A_{2}O_{5}]}{\Delta t}$ = k’ $[A_{2}O_{5}]$

The half-life equation would stand as $t_{1/2}$ = $\frac{0.693}{2k’}$

We have to note that rates of each nothing but the concentration of the reactant at time t = 0 and later at time t. The ratio of concentration is the fraction of reactant that remains after a given time has elapsed.

Natural logarithm $\frac{(conc. \ of \ ‘R’ \ after \ time \ t)}{conc. \ of \ ‘R’ \ at \ the \ start \ of \ reaction)}$ = ln (fraction of ‘R’ remaining at time t) = $\left(\frac{-rate \ constant}{elapsed \ time}\right)$

The negative signs of equation is because the ratio is less than 1. The convenient units for ‘R’ are mostly moles per liter, moles gram, number of atoms, and number of molecules or even gas pressure.

When we have a solution reaction taking place involving an acid and a base we get to observe neutralisation reaction. A strong acid like HCl dissociates completely in water which gives H+ and Cl- ions and similarly a strong base like water soluble metal hydroxide of NaOH dissolves in water and dissociate to give Na+ and OH- ions.

**When the strong acid and a strong base reacts the net ionic reaction is shown as **

An acid base reaction is often called a neutralization reaction as when we react just the exact amount of strong acid and strong base in a solution we observe neutralization and hence the steps that are involved with the stoichiometry part of neutralization reaction follows the same intermediate steps that we get to see in others.

HCl being a very strong acid, it dissociates completely to produce H+ ions and Cl- ions and when the strong base NaOH dissolves, that also dissociates completely. This finally results in net ionic equation for the reaction which is mentioned earlier.

When we calculate the yield of any particular reagent in a chemical change we need to cross check the following things before we arrive at the final yield. Identifying the limiting reagent if any in the reaction. The moles are a different way of expressing the number of atoms, ions, or molecules. Reaction ratios are the key to reaction stoichiometry and once the reaction is balanced we can make a reaction ratio for any reactant and product present in the change.

**The balance reaction between calcium and water is as follows:**

From the above balanced reaction we can see that one mole of calcium reacts with 2 moles of water to yield 1 mole of hydrogen.

The simple ratio that works out here also shows that if 3 moles of calcium is used then 3 moles of hydrogen is expected to yield as well.

**Here are some of the problems based on Reaction Stoichiometry:**

**Question 1:** Methane is combusted in presence of excess of oxygen to produce carbon dioxide and water. What is the mass of carbon dioxide produced if 96 L of methane is combusted under STP?

**Solution:**

$CH_{4} + 2 O_{2} \rightarrow CO_{2} + 2 H_{2}O$

From the balanced reaction it is clear that the ratio between methane and carbon dioxide is 1:1

96 L of methane under STP coverts into $\frac{96}{22.4}$ = 4.285 moles

The moles of methane used for the reaction is 4.285 and hence the same number of moles will be utilised for carbon di oxide as well.

Mass of one mole of carbon dioxide is 44 g.

Therefore, 4.285 moles would give (4.285 $\times$ 44.0) =**188.57 g (approx.) of carbon dioxide**.

**Question 2: **During the formation of water 4.7 litres of oxygen reacts with 7.5 litres of hydrogen gas to produce 2.2 litres of water. What is the percent yield for water and which one of the reagent is the limiting factor?

**Solution:**

The balanced reaction of hydrogen and oxygen forming water is as follows:

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

From the above given we could see the oxygen mole ratio as compared to hydrogen is 1:2

Therefore, the volume over mole ratio of oxygen and hydrogen would show

Oxygen $\frac{V}{n}$ : $\frac{4.7}{1}$ = 4.7

Hydrogen $\frac{V}{n}$ : $\frac{7.5}{2}$ = 3.75

The volume of oxygen which will remain unreacted is 4.7 – 3.75 = 0.95 L

The limiting reagent is hydrogen which gets over first.

The molar ratio between hydrogen and water being same the volume of water that will form is same as hydrogen used up. Hence the volume of water is 3.75 L

Yield percent = $\frac{2.2}{3.75}$ $\times$ 100 =**58.67 %**

Each individual steps is called

The amount of products calculated using limiting reactant concept is considered as theoretical yield. The amount of products which are experimentally formed in a reaction is considered to be the actual yield. There is a difference between actual and theoretical yield as the chemical reactions do not always produce the maximum possible amount of products.

The percent yield formula rule is used to determine the theoretical yield of a particular reaction.

There are many chemical reaction processes where a single overall reaction is never found to suffix the chemical change. The stoichiometry of oxidation of hydrocarbons sourced from exhaust gases in the troposphere cannot be described by a single reaction and many of the hydrocarbons emitted to troposphere and the change in ratio dependent on the type of pollution source.

Qualitatively the rate of reaction is easy way to understand as it represent the change in concentration of the reactants as well as products. When we deal with rates of reaction quantitatively we need specific

The numerical values of the rate of formation of reactants and products are related but different due to reaction stoichiometry. The reaction for rate is given out by dividing each rate by its stoichiometry coefficient.

If we have a reaction of this nature A is decomposing into two compounds then we have a working relation of the reaction as:

$2 A_{2} O_{5} \rightarrow 4 BO_{2} + C_{2}$

Rate = $\frac{\Delta [C_{2}]}{\Delta t}$ = $\frac{1}{2}$ $\frac{[\Delta A_{2}O_{5}]}{\Delta t}$ = $\frac{1}{4}$ $\frac{[BO_{2}]}{\Delta t}$

There are different information in the equation mentioned above. It gives the relation between the rates of change of concentration of reactants and products.

Based on

Rate = $\frac{1}{2}$ $\frac{[\Delta A_{2}O_{5}]}{\Delta t}$ = k $A_{2}O_{5}$

Or, rate = $\frac{[\Delta A_{2}O_{5}]}{\Delta t}$ = 2k $A_{2}O_{5}$

The integrated equation should look like: ln $\frac{[A_{2}O_{5}]t}{[A_{2}O_{5}]0}$ = -2kt

Hence the half-life equation would be $t_{1/2}$ = $\frac{0.693}{2k}$

- $\frac{\Delta [A_{2}O_{5}]}{\Delta t}$ = k’ $[A_{2}O_{5}]$

The half-life equation would stand as $t_{1/2}$ = $\frac{0.693}{2k’}$

We have to note that rates of each nothing but the concentration of the reactant at time t = 0 and later at time t. The ratio of concentration is the fraction of reactant that remains after a given time has elapsed.

Natural logarithm $\frac{(conc. \ of \ ‘R’ \ after \ time \ t)}{conc. \ of \ ‘R’ \ at \ the \ start \ of \ reaction)}$ = ln (fraction of ‘R’ remaining at time t) = $\left(\frac{-rate \ constant}{elapsed \ time}\right)$

The negative signs of equation is because the ratio is less than 1. The convenient units for ‘R’ are mostly moles per liter, moles gram, number of atoms, and number of molecules or even gas pressure.

When we have a solution reaction taking place involving an acid and a base we get to observe neutralisation reaction. A strong acid like HCl dissociates completely in water which gives H+ and Cl- ions and similarly a strong base like water soluble metal hydroxide of NaOH dissolves in water and dissociate to give Na+ and OH- ions.

$H^{+} (aq) + OH^{-} (aq) \rightarrow H_{2}O(l)$

HCl being a very strong acid, it dissociates completely to produce H+ ions and Cl- ions and when the strong base NaOH dissolves, that also dissociates completely. This finally results in net ionic equation for the reaction which is mentioned earlier.

When we calculate the yield of any particular reagent in a chemical change we need to cross check the following things before we arrive at the final yield. Identifying the limiting reagent if any in the reaction. The moles are a different way of expressing the number of atoms, ions, or molecules. Reaction ratios are the key to reaction stoichiometry and once the reaction is balanced we can make a reaction ratio for any reactant and product present in the change.

- Convert the mass to moles
- Reaction rates if any, needs to be identified
- Converting moles back into mass
- Determine the limiting reagent
- Finally determine the maximum amount of product possible

$Ca + 2 H_{2}O$ = $H_{2} + Ca (OH)_{2}$

The simple ratio that works out here also shows that if 3 moles of calcium is used then 3 moles of hydrogen is expected to yield as well.

$3 Ca + 6 H_{2}O$ = $3 H_{2} + 3 Ca (OH)_{2}$

$CH_{4} + 2 O_{2} \rightarrow CO_{2} + 2 H_{2}O$

From the balanced reaction it is clear that the ratio between methane and carbon dioxide is 1:1

96 L of methane under STP coverts into $\frac{96}{22.4}$ = 4.285 moles

The moles of methane used for the reaction is 4.285 and hence the same number of moles will be utilised for carbon di oxide as well.

Mass of one mole of carbon dioxide is 44 g.

Therefore, 4.285 moles would give (4.285 $\times$ 44.0) =

The balanced reaction of hydrogen and oxygen forming water is as follows:

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

From the above given we could see the oxygen mole ratio as compared to hydrogen is 1:2

Therefore, the volume over mole ratio of oxygen and hydrogen would show

Oxygen $\frac{V}{n}$ : $\frac{4.7}{1}$ = 4.7

Hydrogen $\frac{V}{n}$ : $\frac{7.5}{2}$ = 3.75

The volume of oxygen which will remain unreacted is 4.7 – 3.75 = 0.95 L

The limiting reagent is hydrogen which gets over first.

The molar ratio between hydrogen and water being same the volume of water that will form is same as hydrogen used up. Hence the volume of water is 3.75 L

Yield percent = $\frac{2.2}{3.75}$ $\times$ 100 =

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Theoretical Yield | Actual Yield |

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