NaClO3 is used, even in spacecrafts, to produce O2. The daily consumption of pure O2 by a person is

Q: Why do we use the Ideal Gas Law here instead of just using 22.4 L/mol?

The value of 22.4 L/mol is only valid at Standard Temperature and Pressure (STP), which is 273.15 K and 1 atm. Here, the temperature is 300 K, so the gas has expanded. We must use the full Ideal Gas Law ($PV = nRT$) to find the correct number of moles.

Q: How did we know the stoichiometric ratio is 1:1?

By looking at the coefficients in the balanced chemical equation: $\text{NaClO}_3(s) + \text{Fe}(s) \rightarrow \text{O}_2(g) + \text{NaCl}(s) + \text{FeO}(s)$. Since there are no numbers written in front of $\text{NaClO}_3$ and $\text{O}_2$, it implies a coefficient of 1 for both.

Q: What if the temperature was given in Celsius?

You must always convert temperature to Kelvin when using the Ideal Gas Law. You do this by adding 273.15 (or just 273 for simplicity if not specified) to the Celsius value.

Q: Why is the mass of Fe not considered in the calculation?

Iron (Fe) is indeed a reactant, but the question specifically asks for the amount of $\text{NaClO}_3$ required to produce the given amount of $\text{O}_2$. We only need to relate the specific reactant asked about to the product given.

Q: How do I calculate the molar mass of $\text{NaClO}_3$?

You add the atomic masses of its constituent elements: Na (23) + Cl (35.5) + 3 * O (16) = 106.5 g/mol. Always remember to multiply the mass of oxygen by 3 because of the subscript.

Question

$	ext{NaClO}_3$ is used, even in spacecrafts, to produce $	ext{O}_2$. The daily consumption of pure $	ext{O}_2$ by a person is $492	ext{ L}$ at $1	ext{ atm}$, $300	ext{ K}$. How much amount of $	ext{NaClO}_3$, in grams, is required to produce $	ext{O}_2$ for the daily consumption of a person at $1	ext{ atm}$, $300	ext{ K}$ .......... ?
$	ext{NaClO}_3(s) + 	ext{Fe}(s) ightarrow 	ext{O}_2(g) + 	ext{NaCl}(s) + 	ext{FeO}(s)$
$R = 0.082	ext{ L atm mol}^{-1}	ext{K}^{-1}$

Sigma Percentile · Accepted Answer

### The Spacecraft Challenge

Imagine you are an engineer tasked with designing the life support system for a deep-space mission. Your astronaut requires a steady, reliable supply of oxygen to survive. Instead of carrying heavy, pressurized oxygen tanks, you decide to use a chemical oxygen generator—often called a "chlorate candle."

The reaction relies on sodium chlorate ($	ext{NaClO}_3$) reacting with iron ($	ext{Fe}$) to produce oxygen gas ($	ext{O}_2$). The question asks us to determine exactly how much sodium chlorate is needed to meet the daily oxygen consumption of one astronaut, given the specific conditions of the spacecraft cabin.

### Decoding the Oxygen Demand

Before we can weigh out any chemicals, we need to know exactly how much oxygen the astronaut breathes in terms of molecules, or **moles**. We are given the macroscopic properties of the gas:
- Volume ($V$) = $492	ext{ L}$
- Pressure ($P$) = $1	ext{ atm}$
- Temperature ($T$) = $300	ext{ K}$

Since oxygen behaves very much like an ideal gas under these conditions, the **Ideal Gas Law** is our perfect tool:
$$PV = nRT$$

We can rearrange this to solve for the number of moles ($n$):
$$n_{	ext{O}_2} = \frac{PV}{RT}$$

Substituting the given values, along with the universal gas constant $R = 0.082	ext{ L atm K}^{-1}	ext{mol}^{-1}$:
$$n_{	ext{O}_2} = \frac{1 	imes 492}{0.082 	imes 300}$$

Calculating the denominator first, $0.082 	imes 300 = 24.6$. Now, dividing the volume by this result:
$$n_{	ext{O}_2} = \frac{492}{24.6} = 20	ext{ mol}$$

Our astronaut requires exactly **$20	ext{ moles}$** of oxygen gas every single day.

### The Stoichiometric Bridge

Now that we know the target amount of oxygen, we must look at the chemical reaction that produces it:
$$	ext{NaClO}_3(s) + 	ext{Fe}(s) ightarrow 	ext{O}_2(g) + 	ext{NaCl}(s) + 	ext{FeO}(s)$$

This balanced equation is the bridge between the product we want and the reactant we need. Notice the coefficients (the numbers in front of the molecules). There is an invisible "1" in front of both $	ext{NaClO}_3$ and $	ext{O}_2$.

This tells us that the stoichiometric ratio is **1:1**. For every one mole of oxygen gas produced, exactly one mole of sodium chlorate must be consumed. Therefore, to produce $20	ext{ moles}$ of $	ext{O}_2$, we must start with exactly $20	ext{ moles}$ of $	ext{NaClO}_3$.

### The Final Payload

We can't measure out "moles" on a scale; we need to convert this amount into grams. To do this, we calculate the **molar mass** of sodium chlorate by summing the atomic masses of its constituent elements:
- Sodium ($	ext{Na}$) = $23	ext{ g/mol}$
- Chlorine ($	ext{Cl}$) = $35.5	ext{ g/mol}$
- Oxygen ($	ext{O}$) = $16	ext{ g/mol} 	imes 3 = 48	ext{ g/mol}$

$$M_{	ext{NaClO}_3} = 23 + 35.5 + 48 = 106.5	ext{ g/mol}$$

Finally, we multiply the required number of moles by the molar mass to find the total mass needed:
$$m_{	ext{NaClO}_3} = n 	imes M$$
$$m_{	ext{NaClO}_3} = 20	ext{ mol} 	imes 106.5	ext{ g/mol} = 2130	ext{ g}$$

To keep the astronaut breathing for one day, the life support system must consume exactly **$2130	ext{ grams}$** of sodium chlorate. Mission accomplished!

Enter Numerical Value:

Visualized Solution

$The Spacecraft Setup$

$Ideal Gas Law$

$Substituting Gas Parameters$

$Calculating Moles of O_{2}$

$Reaction Stoichiometry$

$Moles of NaClO_{3}$

$Molar Mass of NaClO_{3}$

$Calculating Final Mass$

$What’s Next?$

The Sigma Insight: Stoichiometric and Volumetric Calculations

The Spacecraft Challenge

Decoding the Oxygen Demand

The Stoichiometric Bridge

The Final Payload

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Spacecraft Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ideal Gas Law

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Gas Parameters

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Moles of O2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Reaction Stoichiometry

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Moles of NaClO3​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Molar Mass of NaClO3​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Final Mass

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } What’s Next?