Question

Given the balanced equation \(4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\), how many grams of \(\mathrm{NH}_{3}\) will be required to react with \(80 \mathrm{~g}\) of \(\mathrm{O}_{2}\) ?

Short answer

The required mass of ammonia (NH3) to react completely with 80 g of oxygen (O2) is \(34.08 \text{ g}\).

Step-by-step solution

Write down the given balanced equation

We start by writing down the given balanced equation : \(4NH_3 (g) + 5O_2 (g) \rightarrow 4NO (g) + 6H_2O (g)\)

Calculate the molar mass of ammonia (NH3) and oxygen (O2)

We determine the molar mass of NH3 and O2 by adding up the atomic masses of their respective elements (N, H, and O): Molar mass of NH3 = \(1 \times 14.01 (\text{N}) + 3 \times 1.01 (\text{H}) = 17.04 \text{ g/mol}\) Molar mass of O2 = \(2 \times 16.00 (\text{O}) = 32.00 \text{ g/mol}\)

Calculate the amount of moles of O2 given in the problem

To find the number of moles of O2, we divide the given mass of O2 by its molar mass: moles of O2 = \( \frac{80 \text{g}}{32.00 \text{ g/mol}} = 2.50 \text{ moles}\)

Use stoichiometry to find the required moles of NH3

According to the balanced equation, the stoichiometric ratio between NH3 and O2 is 4:5, which means that 4 moles of NH3 are required for every 5 moles of O2. We use this ratio to find the required moles of NH3: moles of NH3 = \( \frac{4 \text{ moles NH}_3}{5 \text{ moles } O_2} \times 2.50 \text{ moles } O_2 = 2.00 \text{ moles NH}_3\)

Calculate the required mass of NH3

Finally, we calculate the mass of NH3 required by multiplying its moles by its molar mass: mass of NH3 = \(2.00 \text{ moles NH}_3 \times 17.04 \text{ g/mol} = 34.08 \text{ g}\) The required mass of ammonia (NH3) to react completely with 80 g of oxygen (O2) is 34.08 g.

Key concepts

Balanced Chemical Equation

Understanding a balanced chemical equation is fundamental in solving stoichiometry problems. A chemical equation represents the reactants and products in a chemical reaction. For the equation to be balanced, the number of atoms for each element must be the same on both sides of the equation. Taking our given problem, the equation is balanced as follows: \(4 \text{NH}_3 + 5 \text{O}_2 \rightarrow 4 \text{NO} + 6 \text{H}_2\text{O}\).Four ammonia molecules react with five oxygen molecules to yield four nitric oxide molecules and six water molecules. The numbers in front of each molecule are known as stoichiometric coefficients and play a crucial role in calculating quantities of reactants and products during a chemical reaction.

Molar Mass Calculation

The molar mass is the weight in grams of one mole of a substance. It is essential for converting between moles and grams. To calculate molar mass, sum up the atomic masses (from the periodic table) of all atoms in the molecule. For instance:

• Molar mass of NH3 = \(1 \times 14.01 (N) + 3 \times 1.01 (H) = 17.04 \text{ g/mol}\)
• Molar mass of O2 = \(2 \times 16.00 (O) = 32.00 \text{ g/mol}\)

Each hydrogen (H) has an approximate atomic mass of 1.01 g/mol, nitrogen (N) has 14.01 g/mol, and oxygen (O) has 16.00 g/mol.

Stoichiometric Ratios

Stoichiometric ratios stem from a balanced chemical equation and guide the proportions in which reactants combine and determine how much of each product forms. They are ratio of moles of one substance to the moles of another substance in a reaction. In our exercise, the stoichiometric ratio of NH3 to O2 is 4:5. This indicates that for every 5 moles of O2 consumed, 4 moles of NH3 are required. It can be used to calculate the amount of reactants needed or product formed in a reaction, as seen in the steps provided.

Moles to Grams Conversion

Moles to grams conversion is key in linking the theoretical (molar) quantity with a practical (mass) measurement. This is done by multiplying the number of moles by the substance's molar mass. To find the mass of NH3 needed to react with 80 g of O2:

• First, convert the mass of O2 to moles by dividing by its molar mass.
• Use the stoichiometric ratios to find how many moles of NH3 are needed.
• Then, convert moles of NH3 back to grams by multiplying by its molar mass: \(2.00 \text{ moles NH}_3 \times 17.04 \text{ g/mol} = 34.08 \text{ g}\).

This conversion is a two-way street—you can convert from moles to grams or grams to moles depending on the direction of your calculation and what is given or sought in the problem.