To find the volume of ammonia gas (NH₃) that will form at standard temperature and pressure (STP) from 80.25 grams of ammonium chloride (NH₄Cl), follow these steps:
1. Calculate the moles of NH₄Cl:
- First, determine the molar mass of NH₄Cl. The molar masses of its components are approximately:
- Nitrogen (N): 14.01 g/mol
- Hydrogen (H): 1.01 g/mol, but there are 4 hydrogen atoms, so 4 * 1.01 = 4.04 g/mol
- Chlorine (Cl): 35.44 g/mol
- Adding these up, the molar mass of NH₄Cl is 14.01 + 4.04 + 35.44 = 53.49 g/mol
- Use the given mass of NH₄Cl (80.25 g) and the molar mass of NH₄Cl to calculate the moles:
[tex]\[
\text{Moles of NH₄Cl} = \frac{\text{mass of NH₄Cl}}{\text{molar mass of NH₄Cl}} = \frac{80.25 \text{ g}}{53.49 \text{ g/mol}} \approx 1.5003 \text{ moles}
\][/tex]
2. Interpret the balanced chemical reaction:
- The decomposition of NH₄Cl is represented by the balanced equation:
[tex]\[
\text{NH₄Cl} (s) \rightarrow \text{NH₃} (g) + \text{HCl} (g)
\][/tex]
- From the equation, we see that 1 mole of NH₄Cl produces 1 mole of NH₃.
3. Determine the volume of NH₃ gas produced:
- At standard temperature and pressure (STP), 1 mole of any ideal gas occupies 22.414 liters.
- Using the moles of NH₄Cl, which is equal to the moles of NH₃ produced:
[tex]\[
\text{Volume of NH₃ gas} = \text{moles of NH₃} \times \text{molar volume at STP} = 1.5003 \text{ moles} \times 22.414 \text{ L/mol} \approx 33.63 \text{ L}
\][/tex]
Therefore, the volume of ammonia gas (NH₃) produced at STP from 80.25 grams of NH₄Cl is approximately 33.63 liters.
