The balanced chemical reaction given is:

$\text{Mn(s)} + \text{O}_2(\text{g}) \rightarrow \text{MnO}_3(\text{s})$

We need to determine how much product, manganese(VI) oxide ($\text{MnO}_3$), is produced when manganese (Mn) completely reacts with oxygen ($\text{O}_2$).

Step 1: Balance the chemical equation.

The reaction as written is not balanced. To balance it, let's first examine the number of atoms of each element on both sides.

1. Mn atoms: There is 1 Mn atom on both the left and right sides.
2. O atoms: On the left side, we have $\text{O}_2$, which has 2 oxygen atoms. On the right side, we have $\text{MnO}_3$, which contains 3 oxygen atoms.

To balance the oxygen atoms, we can place a coefficient of 3/2 (or 1.5) in front of $\text{O}_2$ on the left side, but this creates fractional coefficients. To avoid fractions, we multiply the entire equation by 2:

$2 \text{Mn(s)} + 3 \text{O}_2(\text{g}) \rightarrow 2 \text{MnO}_3(\text{s})$

Now the equation is balanced, with 2 Mn atoms and 6 O atoms on both sides.

Step 2: Relating Reactants to Products.

From the balanced equation:

• 2 moles of Mn react with 3 moles of $\text{O}_2$ to produce 2 moles of $\text{MnO}_3$.

If you know the quantity of Mn or $\text{O}_2$ you start with, you can use stoichiometry to find out how much $\text{MnO}_3$ is produced.

Would you like to provide the amount of Mn or $\text{O}_2$, or should I proceed with an example calculation?

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Related Questions:

1. How do you determine the molar mass of MnO₃?
2. What is the significance of balancing a chemical equation?
3. How do you perform stoichiometric conversions in a chemical reaction?
4. Can you calculate the mass of MnO₃ produced if a certain mass of Mn reacts?
5. What are the steps to calculate the limiting reagent in a chemical reaction?

Tip: Always balance chemical equations before performing any stoichiometric calculations.