To find the limiting reactant, we need to first write the balanced chemical equation for the reaction between nitrogen dioxide ($NO_2$) and water ($H_2O$):

$3 NO_2 (g) + H_2O (l) \rightarrow 2 HNO_3 (aq) + NO (g)$

From the balanced equation, we can see that 3 moles of $NO_2$ react with 1 mole of $H_2O$ to form the products.

Now, we will compare the mole ratio of the reactants you have with the ratio required by the balanced equation:

• You have 7 moles of $NO_2$.
• You have 11 moles of $H_2O$.

Step 1: Calculate the mole ratio based on the available moles

• According to the equation, 3 moles of $NO_2$ react with 1 mole of $H_2O$. So for 7 moles of $NO_2$, the required moles of $H_2O$ would be:

$\frac{7}{3} = 2.33 \text{ moles of } H_2O \text{ needed}.$

Step 2: Compare with the available moles of $H_2O$

• You have 11 moles of $H_2O$, which is much more than the required 2.33 moles.

Therefore, $NO_2$ is the limiting reactant because you do not have enough of it to react with all the $H_2O$.

The limiting reactant is:

$NO_2$

Would you like to go over the process in more detail, or do you have any questions?

Here are 5 related questions to expand your understanding:

1. How do you determine the excess reactant in a chemical reaction?
2. What is the concept of the limiting reactant in stoichiometry?
3. How do you calculate the theoretical yield of a product based on the limiting reactant?
4. Why is balancing a chemical equation important for stoichiometry problems?
5. What happens when a limiting reactant is completely consumed in a reaction?

Tip: Always double-check the mole ratio from the balanced equation before determining the limiting reactant.