Concept:Convert the mass of the limiting reactant (Carbon) into moles. Because the reaction has a 1:1 molar ratio, the moles of product (\( \text{CO}_2 \)) will be identical, which can then be converted into the number of molecules.
Formula:$$ \text{Molecules} = \left( \frac{m}{M} \right) \times N_A $$
Solution:- Mass of Carbon = 3.0 g. Molar mass of Carbon = 12 g/mol.
- Moles of Carbon = \( 3.0 / 12 = 0.25 \text{ moles} \) (or \( \frac{1}{4} \text{ mole} \)).
- From the equation \( \text{C} + \text{O}_2 \rightarrow \text{CO}_2 \), 1 mole of C produces 1 mole of \( \text{CO}_2 \).
- Therefore, 0.25 moles of C produces exactly 0.25 moles of \( \text{CO}_2 \).
- Calculate the number of molecules: \( 0.25 \times (6.022 \times 10^{23}) \).
- \( \frac{6.022}{4} \approx 1.505 \). Thus, there are \( 1.505 \times 10^{23} \text{ molecules} \).
Why other options are incorrect:- 5.5g & 22g: 0.25 moles of \( \text{CO}_2 \) weighs 11g (not 5.5g or 22g).
- \( 6.02 \times 10^{23} \): This is the number of molecules in a full 1.0 mole (which would require 12g of Carbon).
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