2NaClO_(3)(s)arrow 2NaCl(s)+3O_(2)(g) 1200 moles of NaClO_(3) will produce how many grams of O_(2) 256 g of O_(2) 567g of O_(2) 288 g of O_(2) 32 got O_(2)

To solve this problem, we need to use the stoichiometry of the balanced chemical equation and the molar mass of oxygen gas (O2).<br /><br />Given information:<br />- Balanced chemical equation: $2NaClO_{3}(s)\rightarrow 2NaCl(s)+3O_{2}(g)$<br />- 1200 moles of $NaClO_{3}$<br /><br />Step 1: Determine the moles of $O_{2}$ produced.<br />According to the balanced equation, 2 moles of $NaClO_{3}$ produce 3 moles of $O_{2}$.<br />Therefore, 1200 moles of $NaClO_{3}$ will produce:<br />$\frac{3}{2} \times 1200 = 1800$ moles of $O_{2}$<br /><br />Step 2: Calculate the mass of $O_{2}$ produced.<br />The molar mass of $O_{2}$ is 32 g/mol.<br />Therefore, the mass of $O_{2}$ produced is:<br />$1800 \text{ moles} \times 32 \text{ g/mol} = 57600 \text{ g}$<br /><br />Therefore, 1200 moles of $NaClO_{3}$ will produce 57600 grams of $O_{2}$.<br /><br />The correct answer is: 567g of $O_{2}$.