Animated Solution for Chemistry - Basic Concepts in Chemistry: The minimum amount of ${\text{O}}_2(g)$ consumed per gram of reactant is for the reaction (Given atomic mass : $\text{Fe}=56,\text{O}=16,\text{Mg}=24,\text{P}=31,\text{C}=12,\text{H}=1$)

Imagine you are a chemical engineer tasked with designing a life-support system for a spacecraft. You have a limited supply of oxygen, and you need to choose a reaction that consumes the absolute minimum amount of oxygen per gram of fuel (reactant) you carry. This is exactly what this problem is asking us to find!

We are given four different chemical reactions, and our goal is to determine which one is the most "oxygen-efficient" on a per-gram basis. To do this, we need to bridge the gap between the microscopic world of atoms and the macroscopic world of grams using the magic of stoichiometry.

The balanced chemical equations are our recipes. They tell us exactly how many moles of each ingredient we need. However, the question asks for the amount of oxygen consumed per gram of reactant, not per mole.

This means we need to calculate two things for each reaction: 1. The total mass of the reactant involved in the balanced equation. 2. The total mass of oxygen gas ($O_2$) consumed in that same equation.

Once we have these two masses, we simply divide the mass of oxygen by the mass of the reactant to find our ratio. Let's crunch the numbers!

Reaction A: The Combustion of Propane Our first candidate is propane ($C_3{H}_8$). The balanced equation is:

One mole of propane reacts with five moles of oxygen. The molar mass of propane is $(3\times 12)+(8\times 1)=44\text{ g/mol}$. The mass of five moles of oxygen is $5\times 32=160\text{ g}$.

The ratio is $\frac{160}{44}\approx 3.64\text{ g}$ of oxygen per gram of propane. That's a lot of oxygen! Hydrocarbons are notoriously oxygen-hungry.

Reaction B: The Oxidation of Phosphorus Next, we have solid phosphorus ($P_4$):

One mole of $P_4$ reacts with five moles of oxygen. The molar mass of $P_4$ is $4\times 31=124\text{ g/mol}$. The mass of five moles of oxygen is again $160\text{ g}$.

The ratio is $\frac{160}{124}\approx 1.29\text{ g}$ of oxygen per gram of phosphorus. Much better than propane, but can we go lower?

Reaction C: The Rusting of Iron Let's look at iron ($Fe$):

Four moles of iron react with three moles of oxygen. The mass of four moles of iron is $4\times 56=224\text{ g}$. The mass of three moles of oxygen is $3\times 32=96\text{ g}$.

The ratio is $\frac{96}{224}\approx 0.43\text{ g}$ of oxygen per gram of iron. This is incredibly low! Iron is very heavy, so a gram of iron doesn't contain that many atoms compared to a gram of carbon or hydrogen.

Reaction D: The Combustion of Magnesium Finally, magnesium ($Mg$):

Two moles of magnesium react with one mole of oxygen. The mass of two moles of magnesium is $2\times 24=48\text{ g}$. The mass of one mole of oxygen is $32\text{ g}$.

The ratio is $\frac{32}{48}\approx 0.67\text{ g}$ of oxygen per gram of magnesium.

Comparing our results, we have: - Propane: $3.64\text{ g}$ - Phosphorus: $1.29\text{ g}$ - Iron: $0.43\text{ g}$ - Magnesium: $0.67\text{ g}$

The clear winner is iron! It consumes the absolute minimum amount of oxygen per gram of reactant. This makes sense intuitively: iron has the highest atomic mass among the elements listed, meaning one gram of iron contains fewer atoms to react with oxygen compared to a gram of the lighter elements.

Always remember, in stoichiometry, the coefficients tell you the mole ratio, but the molar masses dictate the mass ratio. Don't let the coefficients fool you!