During lightning flashes, nitrogen combines with oxygen in the atmosphere to form nitrogen monoxide, NO, which then reacts further with \(\mathrm{O}_{2}\) to produce nitrogen dioxide, NO. a. What mass of \(\mathrm{NO}_{2}\) , is formed when NO reacts with 384 \(\mathrm{g} \mathrm{O}_{2} ?\) b. How many grams of NO are required to react with this amount of \(\mathrm{O}_{2} ?\)

Understanding balanced chemical equations is essential when solving stoichiometry problems. A balanced chemical equation shows the relationship between reactants and products. In this exercise, we start by balancing the equations for the reactions:

For the formation of NO: \ \[\text{N}_{2} + \text{O}_{2} \rightarrow 2\text{NO}\]\

This equation indicates that one molecule of nitrogen gas (\(\text{N}_{2}\)) reacts with one molecule of oxygen gas (\(\text{O}_{2}\)) to produce two molecules of nitrogen monoxide (\(\text{NO}\)). This is a balanced reaction because there are equal numbers of each type of atom on both sides of the equation.

Next, in the formation of nitrogen dioxide (\(\text{NO}_{2}\)):

\[2\text{NO} + \text{O}_{2} \rightarrow 2\text{NO}_{2}\]\

This equation tells us that two molecules of nitrogen monoxide react with one molecule of oxygen gas to produce two molecules of nitrogen dioxide. Again, the equation is balanced because the count of nitrogen and oxygen atoms is the same on both sides. These balanced equations are crucial because they help us determine the exact amounts of substances involved in the reactions.

Molar mass calculations are the next step in solving stoichiometry problems. The molar mass of a substance is the mass of one mole of that substance, typically expressed in grams per mole (g/mol). Molar masses are determined from the atomic masses of the elements involved.

For example, to calculate the molar mass of \(\text{O}_{2}\), the diatomic molecule of oxygen, you add the atomic masses of two oxygen atoms:

\[\text{Molar mass of } \text{O}_{2} = 2 \times 16 \text{ g/mol} = 32 \text{ g/mol}\]

For \(\text{NO}\), the molar mass is calculated by adding the atomic masses of nitrogen (N) and oxygen (O):

\[\text{Molar mass of } \text{NO} = 14 (\text{N}) + 16 (\text{O}) = 30 \text{ g/mol}\]

And for \(\text{NO}_{2}\), we incorporate one nitrogen and two oxygen atoms:

\[\text{Molar mass of } \text{NO}_{2} = 14 (\text{N}) + 2 \times 16 (\text{O}) = 46 \text{ g/mol}\]

Understanding how to calculate molar masses allows you to convert between grams and moles, which is necessary for solving stoichiometry problems.

Mole-to-mass conversions are pivotal in determining the quantities of substances involved in chemical reactions. This involves using the molar mass of a substance to convert between the mass of a substance and the amount of substance in moles.

For instance, in our exercise, we first convert the mass of \(\text{O}_{2}\) to moles using its molar mass:

\[\text{Moles of } \text{O}_{2} = \frac{384 \text{ g}}{32 \text{ g/mol}} = 12 \text{ moles}\]

Next, using stoichiometry, we find the moles of \(\text{NO}\) required based on the balanced chemical equation:

\[2 \text{NO} + \text{O}_{2} \rightarrow 2 \text{NO}_{2}\]\

From the reaction, 1 mole of \(\text{O}_{2}\) reacts with 2 moles of \(\text{NO}\). Therefore, 12 moles of \(\text{O}_{2}\) will react with:

\[2 \times 12 = 24 \text{ moles of } \text{NO}\]

Converting the moles of \(\text{NO}\) to mass:

\[\text{Mass of } \text{NO} = 24 \text{ moles} \times 30 \text{ g/mol} = 720 \text{ g}\]

Similarly, we can find the mass of \(\text{NO}_{2}\) formed by first determining the moles of \(\text{NO}_{2}\):

\[\text{Moles of } \text{NO}_{2} = 24 \text{ moles}\]

And then converting these moles to mass:

\[\text{Mass of } \text{NO}_{2} = 24 \text{ moles} \times 46 \text{ g/mol} = 1104 \text{ g}\]

These conversions are fundamental skills in chemistry, enabling us to relate the mass of a substance to the amount present in a reaction.