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Gas Stoichiometry

The quantitative study of the relative amounts of reactants and products in chemical reactions is referred to as stoichiometry. Likewise, quantitative study of the relative amounts of reactants and products, in the chemical reactions that produce gases only, is referred to as gas stoichiometry. The law of conservation of mass is key concept of stoichiometry studies.

According to the law of conservation of mass, the mass of the reactants must be equal to the mass of the products. The same concept is used to calculate the unknown quantities of reactants or products. 

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Gas Stoichiometry Definition

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Gas stoichiometry definition is given as quantitative relationship of reactants and products of a chemical reaction wherein all the products are gases. Studies of gas stoichiometry aim to calculate either the amounts of reactants required for particular amounts of products, or quantities of products produced from particular quantities of reactants. Since gas stoichiometry studies the quantitative relationship of reactant and products of chemical reaction with gaseous products; “gas law” is followed for the calculations. According to the “gas law”, at temperature (T) in Kelvin,the volume (V) occupied by n moles of any gas has a pressure (P). The equation for gas law is given as following:
P V = n R T.

Here R is known as the gas constant and the equation is called the ideal gas law or equation of state. The gas law stands true and efficiently interprets the gas behavior under a temperature much higher than the critical temperature and at low pressures. However, gases at low temperature and at high pressure exhibit slight variations which are then mostly corrected using van der Waal's equation.

How to Do Gas Stoichiometry?

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Gas stiochiometry follows the mole concept and law of partial pressure. Mole concept defines mole as exact amount of substance and refers to the number of atoms. Total number of atoms for an element is equals to it corresponding mole. Thus, mole concept follows the number of identities. Dalton proposed law of partial pressure. Law of partial pressure states that partial pressure for a component of gas is same as that of the component itself in the chamber. Further, the total pressure for that component is sum total of all its partial pressure.

Let’s assume that number of moles and partial pressure of the component 1 is $n_1$ and $P_1$ respectively. Likewise, number of moles and partial pressure of the components 2, 3, 4 and so on would be $n_1$, $n_2$, $n_3$ and $P_1$, $P_2$, $P_3$ ….so on. Let’s suppose that multiple components are present in a container. Therefore, the total number of moles is the sum of number of moles of the components.

This can be represented as following equation:
$n_{total}$ = $n_{1} + n_{2} + n_{3} + . . . + n_{n}$.

According to the gas law:
PV = nRT; 
therefore,
n = $\frac{PV}{RT}$ or $\frac{V}{RT}$ P.

Now, let’s apply the law of partial pressure. Since n = $\frac{V}{RT}$ P, according to the law of partial pressure, the number of moles for the component “i “and its corresponding partial pressure would be:

$n_i$ = $\frac{V}{RT}$ $P_i$ as well as $n_{total}$ = $\frac{V}{RT}$ $P_{total}$

Thus, it the total pressure of that component can be given by following equation:
$P_{total}$ = $P_{1} + P_{2} + P_{3} + . . . + P_{n}$.

Using the above mention information, let’s learn how to do gas stoichiomentry. Let’s assume that 250 mL of HCl gas at 350 K and 75 kPa is dissolved in 100 mL of pure water. Now let’s calculate the concentration of HCl gas.

Solution: As we know that “gas law” states that at temperature (T), the volume (V) occupied by n moles of any gas has a pressure (P). The gas equation is given as following:
P V = n R T
Therefore,
n = $\frac{PV}{RT}$.

As we know that R= gas constant. It value is 8.314 kPa L/(K mol). In the question, P = 75 kPa, V = 250 mL and T = 350K. Now let’s put these values in the above equation to calculate the concentration of HCl.

n = $\frac{0.25 \ L \times 100 \ kPa}{8.314 \ k \ Pa \ L/K \ mol \times 350 \ K}$ = 0.0086 mol

Therefore, concentration of HCl in 100 mL pure water would be:

[HCl] = $\frac{0.00860 \ mol}{0.1 \ L}$ = 0.086 mol/L.


Gas Stoichiometry Problems

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Now, let’s do some gas stoichiometry problems:

For example: Let’s consider the following equation: $C(s) + 2H_{2}(g) \rightarrow CH_{4}(g)$. Given the above equation, let’s calculate the volume of hydrogen which is needed to produce 40.0 L of methane?

Solution: In the question, pressure and temperature are not mentioned. Therefore, Avogadro's hypothesis is applied to solve the problem. According to Avogadro's hypothesis, equal volumes of gas at equal temperature and pressure contain equal moles. Since here, the molar ratio between hydrogen-methane is 2:1.

Therefore,
2 moles hydrogen : 1 mole methane = $x$ L hydrogen : 40.0 L methane
$x$ = 80.0 L of hydrogen.

Mole - Volume Gas Stoichiometry Problems

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The ideal gas law serves as bridge between moles and volumes in the Gas Stoichiometry problems.

For example: Look at the following equation for the combustion of methane: CH4 + 2O2 $\rightarrow$ 2H2O + CO2. Now, let’s suppose that 100.0 L of methane at STP are burned. Under these conditions, calculate the volume of carbon dioxide that will be produced at STP. 

Solution: According to the question, chemical reaction is taking place under standard conditions of temperature and pressure i.e. STP. Therefore, the volumes of reactant and products are directly proportional to the moles of reactant used and product produced. This is because, according to the gas equation: PV = nRT.

It can also be written as:
$\frac{s}{V}$ = $\frac{P}{RT}$.

Now, according to the question, the components in the right side of equation are constant which makes the components on the left side also constant. This means that the n:V ratio must also be constant. Therefore the volume is directly proportional to the number of moles of gas. The question says that molar ratio between methane and carbon dioxide is 1:1 and the volume of methane gas is 100 L. Therefore, the volume of carbon dioxide would also be 100 L. The answer is 100.0 L.
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