Question

calculate the mass of sodium chlorate that is used to form 7.5 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)
17
0.23
25
37

Explanation:

Step1: Calculate moles of \( O_2 \)

Moles of \( O_2 = \frac{\text{mass of } O_2}{\text{molar mass of } O_2} = \frac{7.5\ g}{32.00\ g/mol} \approx 0.2344\ mol \)

Step2: Use stoichiometry to find moles of \( NaClO_3 \)

From the balanced equation \( 2NaClO_3(s)
ightarrow 2NaCl(s) + 3O_2(g) \), the mole ratio of \( NaClO_3 \) to \( O_2 \) is \( \frac{2}{3} \).
Moles of \( NaClO_3 = 0.2344\ mol\ O_2 \times \frac{2\ mol\ NaClO_3}{3\ mol\ O_2} \approx 0.1563\ mol \)

Step3: Calculate mass of \( NaClO_3 \)

Mass of \( NaClO_3 = \text{moles of } NaClO_3 \times \text{molar mass of } NaClO_3 = 0.1563\ mol \times 106.44\ g/mol \approx 16.64\ g \approx 17\ g \) (rounded to a reasonable value matching the options)

Answer:

17