Question

calculate the mass of sodium chlorate that is used to form 6.5 g of oxygen given the balanced chemical equation. (the molar mass of o₂ is 32.00 g/mol; naclo₃ is 106.44 g/mol.)
2naclo₃(s) → 2nacl(s) + 3o₂(g)
options: 32, 14, 22, 0.20

Explanation:

Step1: Calculate moles of \( O_2 \)

Moles of \( O_2 = \frac{\text{mass of } O_2}{\text{molar mass of } O_2} = \frac{6.5\ g}{32.00\ g/mol} \approx 0.2031\ mol \)

Step2: Relate moles of \( O_2 \) to \( NaClO_3 \)

From the balanced equation \( 2NaClO_3
ightarrow 3O_2 \), the mole ratio of \( NaClO_3 \) to \( O_2 \) is \( \frac{2}{3} \).
Moles of \( NaClO_3 = 0.2031\ mol\ O_2 \times \frac{2\ mol\ NaClO_3}{3\ mol\ O_2} \approx 0.1354\ mol \)

Step3: Calculate mass of \( NaClO_3 \)

Mass of \( NaClO_3 = \text{moles} \times \text{molar mass} = 0.1354\ mol \times 106.44\ g/mol \approx 14.41\ g \approx 14\ g \)

Answer:

14