A gaseous mixture contains `SO_(3)(g)` and `C_(2)H_(6)(g)` in a 16:15 ratio by mass. The ratio of total number of atoms present in `C_(2)H_(6)(g)` and `SO_(3)(g)` is:

Answer

To solve the problem, we need to find the ratio of the total number of atoms present in the gases \(C_2H_6\) and \(SO_3\) given their mass ratio of 16:15.

### Step-by-Step Solution:

1. **Identify the Molar Masses:**
- The molar mass of \(SO_3\) (sulfur trioxide) is calculated as follows:
- Sulfur (S) = 32 g/mol
- Oxygen (O) = 16 g/mol (3 O atoms)
- Total molar mass of \(SO_3 = 32 + (3 \times 16) = 32 + 48 = 80 \, \text{g/mol}\)
- The molar mass of \(C_2H_6\) (ethane) is calculated as follows:
- Carbon (C) = 12 g/mol (2 C atoms)
- Hydrogen (H) = 1 g/mol (6 H atoms)
- Total molar mass of \(C_2H_6 = (2 \times 12) + (6 \times 1) = 24 + 6 = 30 \, \text{g/mol}\)

2. **Calculate Moles of Each Gas:**
- Let the mass of \(SO_3\) be \(16x\) grams and the mass of \(C_2H_6\) be \(15x\) grams (since they are in a 16:15 ratio).
- Moles of \(SO_3\):
\[ n_{SO_3} = \frac{\text{mass}}{\text{molar mass}} = \frac{16x}{80} = \frac{x}{5} \]
- Moles of \(C_2H_6\):
\[ n_{C_2H_6} = \frac{15x}{30} = \frac{x}{2} \]

3. **Calculate Total Number of Atoms:**
- Each molecule of \(C_2H_6\) contains 8 atoms (2 Carbon + 6 Hydrogen).
- Total atoms in \(C_2H_6\):
\[ \text{Total atoms in } C_2H_6 = 8 \times n_{C_2H_6} = 8 \times \frac{x}{2} = 4x \]
- Each molecule of \(SO_3\) contains 4 atoms (1 Sulfur + 3 Oxygen).
- Total atoms in \(SO_3\):
\[ \text{Total atoms in } SO_3 = 4 \times n_{SO_3} = 4 \times \frac{x}{5} = \frac{4x}{5} \]

4. **Find the Ratio of Total Atoms:**
- The ratio of total atoms in \(C_2H_6\) to total atoms in \(SO_3\) is:
\[ \text{Ratio} = \frac{\text{Total atoms in } C_2H_6}{\text{Total atoms in } SO_3} = \frac{4x}{\frac{4x}{5}} = 4x \times \frac{5}{4x} = 5 \]

### Final Answer:
The ratio of the total number of atoms present in \(C_2H_6\) and \(SO_3\) is \(5:1\).