Sagot :
SOLUTION:
Based on the balanced chemical equation, 2 moles of NaClO₃ is stoichiometrically equivalent to 3 moles of O₂.
The molar masses of NaClO₃ and O₂ are 106.44 g and 32.00 g, respectively.
[tex]\begin{aligned} \text{mass of} \: \text{NaClO}_3 & = \text{6.5 g} \: \text{O}_2 \times \frac{\text{1 mol} \: \text{O}_2}{\text{32.00 g} \: \text{O}_2} \times \frac{\text{2 mol} \: \text{NaClO}_3}{\text{3 mol} \: \text{O}_2} \times \frac{\text{106.44 g} \: \text{NaClO}_3}{\text{1 mol} \: \text{NaClO}_3} \\ & = \boxed{\text{14.4 g}} \end{aligned}[/tex]
Hence, 14.4 g of NaClO₃ must be decomposed to form 6.5 g of O₂.
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