Question

\\2 \text{ n}_2\text{h}_4(g) + \text{n}_2\text{o}_4(g) \
ightarrow 3 \text{ n}_2(g) + 4 \text{ h}_2\text{o}(g)\\

when \\(8.0\text{ g}\\) of \\(\text{n}_2\text{h}_4\\) (\\(32\text{ g mol}^{-1}\\)) and \\(92\text{ g}\\) of \\(\text{n}_2\text{o}_4\\) (\\(92\text{ g mol}^{-1}\\)) are mixed together and react according to the equation above, what is the maximum mass of \\(\text{h}_2\text{o}\\) that can be produced?

a) \\(9.0\text{ g}\\)
b) \\(18\text{ g}\\)
c) \\(36\text{ g}\\)
d) \\(72\text{ g}\\)
e) \\(144\text{ g}\\)

Explanation:

🆕 New Concept Discovered: Limiting Reactant Stoichiometry
Finding which ingredient runs out first

Step 1: Calculate moles of each reactant

To find out how much product can be made, we first need to convert the given masses of our starting materials into moles. Moles tell us the actual number of molecules we are working with.

For hydrazine (\(\text{N}_2\text{H}_4\)):

$$ \text{Moles of }\text{N}_2\text{H}_4 = \frac{8.0\text{ g}}{32\text{ g mol}^{-1}} = 0.25\text{ mol} $$

For dinitrogen tetroxide (\(\text{N}_2\text{O}_4\)):

$$ \text{Moles of }\text{N}_2\text{O}_4 = \frac{92\text{ g}}{92\text{ g mol}^{-1}} = 1.0\text{ mol} $$

Step 2: Identify the limiting reactant

The balanced chemical equation is:

$$ 2\text{ N}_2\text{H}_4(g) + \text{N}_2\text{O}_4(g) ightarrow 3\text{ N}_2(g) + 4\text{ H}_2\text{O}(g) $$

This tells us we need \(2\) moles of \(\text{N}_2\text{H}_4\) for every \(1\) mole of \(\text{N}_2\text{O}_4\).

Let's determine how many moles of \(\text{N}_2\text{O}_4\) are needed to completely react with our \(0.25\text{ mol}\) of \(\text{N}_2\text{H}_4\):

$$ \text{Required }\text{N}_2\text{O}_4 = 0.25\text{ mol }\text{N}_2\text{H}_4 \times \frac{1\text{ mol }\text{N}_2\text{O}_4}{2\text{ mol }\text{N}_2\text{H}_4} = 0.125\text{ mol} $$

Since we have \(1.0\text{ mol}\) of \(\text{N}_2\text{O}_4\) available, which is much more than the \(0.125\text{ mol}\) required, \(\text{N}_2\text{O}_4\) is in excess.

Therefore, \(\text{N}_2\text{H}_4\) is the limiting reactant and will determine the maximum amount of product we can make.

Step 3: Calculate the maximum moles of \(\text{H}_2\text{O}\) produced

Using the mole ratio between the limiting reactant (\(\text{N}_2\text{H}_4\)) and the desired product (\(\text{H}_2\text{O}\)) from the balanced equation:

$$ \frac{4\text{ mol }\text{H}_2\text{O}}{2\text{ mol }\text{N}_2\text{H}_4} = \frac{2\text{ mol }\text{H}_2\text{O}}{1\text{ mol }\text{N}_2\text{H}_4} $$

Now, calculate the moles of \(\text{H}_2\text{O}\):

$$ \text{Moles of }\text{H}_2\text{O} = 0.25\text{ mol }\text{N}_2\text{H}_4 \times \frac{4\text{ mol }\text{H}_2\text{O}}{2\text{ mol }\text{N}_2\text{H}_4} = 0.50\text{ mol} $$

Step 4: Convert moles of \(\text{H}_2\text{O}\) to mass

The molar mass of water (\(\text{H}_2\text{O}\)) is:

$$ \text{Molar mass of }\text{H}_2\text{O} = (2 \times 1.0\text{ g mol}^{-1}) + 16.0\text{ g mol}^{-1} = 18\text{ g mol}^{-1} $$

Now, calculate the mass:

$$ \text{Mass of }\text{H}_2\text{O} = 0.50\text{ mol} \times 18\text{ g mol}^{-1} = 9.0\text{ g} $$

Answer:

A) 9.0 g