Problem 59 - Chemistry The Molecular Science

Question

Nitrogen monoxide is oxidized in air to give brown nitrogen dioxide. $$2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{~g})$$ Starting with \(2.2 \mathrm{~mol} \mathrm{NO},\) calculate how many moles and how many grams of \(\mathrm{O}_{2}\) are required for complete reaction. Calculate what mass of \(\mathrm{NO}_{2}\), in grams, is produced.

Step-by-Step Solution

Answer

1.1 moles (35.2 g) of \(\text{O}_2\) are required, producing 101.222 g of \(\text{NO}_2\).

Step 1: Write the Balanced Equation

The balanced chemical equation for the reaction is given as: \[ 2 \text{NO}(\text{g}) + \text{O}_{2}(\text{g}) \rightarrow 2 \text{NO}_{2}(\text{g}) \] This equation tells us that 2 moles of nitrogen monoxide react with 1 mole of oxygen to produce 2 moles of nitrogen dioxide.

Step 2: Determine Moles of Oxygen Required

According to the balanced chemical equation, 1 mole of \(\text{O}_2\) is required for every 2 moles of \(\text{NO}\). Given \(2.2\) moles of \(\text{NO}\):\[ \frac{1 \text{ mole of } \text{O}_2}{2 \text{ moles of } \text{NO}} = x \text{ moles of } \text{O}_2/2.2 \text{ moles of } \text{NO} \]Solving for \(x\): \[ x = \frac{1}{2} \times 2.2 = 1.1 \text{ moles of } \text{O}_2 \]Thus, 1.1 moles of \(\text{O}_2\) are required.

Step 3: Convert Moles of Oxygen to Grams

The molar mass of \(\text{O}_2\) is approximately \(32.00 \text{ g/mol}\). To find the mass in grams:\[ 1.1 \text{ moles of } \text{O}_2 \times 32.00 \text{ g/mol} = 35.2 \text{ grams of } \text{O}_2 \]So, \(35.2\) grams of oxygen are required.

Step 4: Calculate Moles of Nitrogen Dioxide Produced

According to the balanced chemical equation, \(2\) moles of \(\text{NO}_2\) are produced from \(2\) moles of \(\text{NO}\). Given \(2.2\) moles of \(\text{NO}\): \[ 2.2 \text{ moles of } \text{NO} \rightarrow 2.2 \text{ moles of } \text{NO}_2 \]Thus, 2.2 moles of \(\text{NO}_2\) are produced.

Step 5: Convert Moles of Nitrogen Dioxide to Grams

The molar mass of \(\text{NO}_2\) is approximately \(46.01 \text{ g/mol}\). To convert the moles to grams:\[ 2.2 \text{ moles of } \text{NO}_2 \times 46.01 \text{ g/mol} = 101.222 \text{ grams of } \text{NO}_2 \]Therefore, \(101.222\) grams of \(\text{NO}_2\) are produced.

Key Concepts

Balanced Chemical Equation

A balanced chemical equation is fundamental to understanding how substances interact in a reaction. It precisely illustrates the proportions in which reactants combine and products form. In the reaction between nitrogen monoxide (NO) and oxygen (\(\text{O}_2\)), the equation is:
\[2 \text{NO}(\text{g}) + \text{O}_{2}(\text{g}) \rightarrow 2 \text{NO}_{2}(\text{g})\]The equation tells us several key points:

• 2 moles of NO react with 1 mole of \(\text{O}_2\)
• 2 moles of \(\text{NO}_2\) are produced

Remember! The coefficients (numbers in front) present the stoichiometric ratio, which ensures mass conservation – the mass and number of atoms are the same on both sides of the reaction.

Molar Mass Calculation

Molar mass is an essential concept. It allows you to convert between an amount of substance (moles) and mass (grams). Think of it as a bridge between tiny atoms and their tangible weights. Here’s how you calculate it:
- Find the atomic mass of each element (found on the periodic table).- For molecules, multiply the atomic mass by the number of atoms.- Add these amounts together for the total molar mass.For instance:- The molar mass of \(\text{O}_2\) is about 32.00 g/mol (16 for each oxygen atom).- The molar mass of \(\text{NO}_2\) is around 46.01 g/mol. Calculate it: 14 (N) + 16 x 2 (O).This concept helps in converting moles to grams using the formula:\[ \text{Mass} = \text{Moles} \times \text{Molar Mass}\]

Mole-to-Mole Ratios

Mole-to-mole ratios are crucial in stoichiometry. They derive directly from the coefficients in a balanced chemical equation. These ratios help determine how much of one substance reacts with or produces another.
Consider the equation:
\[ 2 \text{NO} \rightarrow 1 \text{O}_2 \rightarrow 2 \text{NO}_2\]This tells us:

• 2 moles NO react with 1 mole \(\text{O}_2\)
• 2 moles NO produce 2 moles \(\text{NO}_2\)

To use mole-to-mole ratios, set up proportions to solve for unknown quantities. For example, to find the moles of \(\text{O}_2\) needed for 2.2 moles of NO, notice that the ratio is 2:1. Thus:\[\frac{1 \text{ mole of } \text{O}_2}{2 \text{ moles of } \text{NO}} = \frac{x \text{ moles of } \text{O}_2}{2.2 \text{ moles of } \text{NO}}\]This principle guides you through any mole-based calculations involving chemical reactions.

Chemical Reactions in Gases

Chemical reactions involving gases often occur at constant temperature and pressure, allowing us to apply the ideal gas law principles. A key aspect is to remember gases often behave predictably under given conditions.
In gaseous reactions such as\[2 \text{NO}(\text{g}) + \text{O}_{2}(\text{g}) \rightarrow 2 \text{NO}_{2}(\text{g})\]the relationship between gas volumes (under ideal conditions) mirrors the mole ratios from the chemical equation. Hence, if you use 2 moles of NO and 1 mole of \(\text{O}_2\), their molar volumes also follow this 2:1 ratio in ideal gas conditions.

• 1 mole of any gas occupies 22.4 L at standard temperature and pressure (STP).
• Reactions can also be conducted in non-standard conditions, but calculations adjust them accordingly.

Understanding these gaseous principles simplifies predictions about volumes and quantities needed in reactions.