7. How many grams of ammonium nitrate would you need to have 5.35x1034 atoms of nitrogen?

To determine how many grams of ammonium nitrate \((\text{NH}_4\text{NO}_3)\) would contain \(5.35 \times 10^{34}\) nitrogen atoms, we need to follow these steps:

### Step 1: Molar Mass of Ammonium Nitrate
First, calculate the molar mass of ammonium nitrate \((\text{NH}_4\text{NO}_3)\):
\[
\text{NH}_4\text{NO}_3 = \text{N}_2 + 4\text{H} + 3\text{O}
\]
- Nitrogen (N) = 14.01 g/mol (for 2 nitrogen atoms in NH₄NO₃)
- Hydrogen (H) = 1.01 g/mol (4 atoms of H)
- Oxygen (O) = 16.00 g/mol (3 atoms of O)

Molar mass of \(\text{NH}_4\text{NO}_3\):
\[
= 2(14.01) + 4(1.01) + 3(16.00) = 28.02 + 4.04 + 48.00 = 80.06 \, \text{g/mol}
\]

### Step 2: Number of Nitrogen Atoms per Molecule of Ammonium Nitrate
Each molecule of ammonium nitrate contains 2 nitrogen atoms.

### Step 3: Avogadro's Number
One mole of any substance contains \(6.022 \times 10^{23}\) molecules (Avogadro's number). Since each molecule of ammonium nitrate has 2 nitrogen atoms, 1 mole of ammonium nitrate will contain \(2 \times 6.022 \times 10^{23} = 1.2044 \times 10^{24}\) nitrogen atoms.

### Step 4: Moles of Nitrogen Atoms
To find how many moles of nitrogen atoms correspond to \(5.35 \times 10^{34}\) nitrogen atoms:
\[
\text{moles of nitrogen atoms} = \frac{5.35 \times 10^{34}}{1.2044 \times 10^{24}} = 4.44 \times 10^{10} \, \text{moles of nitrogen atoms}
\]

### Step 5: Moles of Ammonium Nitrate
Since each mole of ammonium nitrate contains 2 moles of nitrogen atoms, the moles of ammonium nitrate will be:
\[
\text{moles of ammonium nitrate} = \frac{4.44 \times 10^{10}}{2} = 2.22 \times 10^{10} \, \text{moles}
\]

### Step 6: Mass of Ammonium Nitrate
Finally, calculate the mass of ammonium nitrate required using its molar mass:
\[
\text{mass} = 2.22 \times 10^{10} \, \text{moles} \times 80.06 \, \text{g/mol} = 1.78 \times 10^{12} \, \text{grams}
\]

### Conclusion:
You would need **\(1.78 \times 10^{12}\) grams (or \(1.78 \, \text{million tons})** of ammonium nitrate to have \(5.35 \times 10^{34}\) nitrogen atoms.

Would you like more details, or do you have any questions?

### Related Questions:
1. How is the molar mass of a compound calculated?
2. What is the significance of Avogadro's number in chemistry?
3. How do you convert between moles and atoms?
4. What are some common uses of ammonium nitrate?
5. How can we determine the percentage of nitrogen in ammonium nitrate?

**Tip:** Always check the number of atoms in a molecule to ensure the correct calculation of moles!

How many grams of ammonium nitrate would you need to have 5.35x10^34 atoms of nitrogen?

Learn how to calculate the mass of ammonium nitrate (NH4NO3) required to have 5.35x10^34 nitrogen atoms. This detailed step-by-step solution explains key concepts such as the mole concept, Avogadro's number, and molar mass calculation, suitable for Grades 10-12 chemistry students.

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