a 115.0 - g sample of oxygen was produced by heating 400.0 g of potassium chlorate.\n2kclo₃ → 2kcl +…

a 115.0 - g sample of oxygen was produced by heating 400.0 g of potassium chlorate.\n2kclo₃ → 2kcl + 3o₂\nwhat is the percent yield of oxygen in this chemical reaction?\nuse %yield = \\frac{actual yield}{theoretical yield}×100.\n69.63%\n73.40%\n90.82%\n136.2%

a 115.0 - g sample of oxygen was produced by heating 400.0 g of potassium chlorate.\n2kclo₃ → 2kcl + 3o₂\nwhat is the percent yield of oxygen in this chemical reaction?\nuse %yield = \\frac{actual yield}{theoretical yield}×100.\n69.63%\n73.40%\n90.82%\n136.2%

Answer

Explanation:

Step1: Calculate molar masses

Molar mass of $KClO_3$: $39.1 + 35.5+3\times16 = 122.6$ g/mol. Molar mass of $O_2$: $2\times16 = 32$ g/mol.

Step2: Determine moles of $KClO_3$

Moles of $KClO_3=\frac{400.0\ g}{122.6\ g/mol}\approx3.2626$ mol.

Step3: Calculate moles of $O_2$ from stoichiometry

From the reaction $2KClO_3\rightarrow2KCl + 3O_2$, moles of $O_2$ produced theoretically $=\frac{3}{2}\times3.2626$ mol $\approx4.8939$ mol.

Step4: Calculate theoretical mass of $O_2$

Theoretical mass of $O_2 = 4.8939$ mol $\times32$ g/mol $=156.6048$ g.

Step5: Calculate percent - yield

Percent - yield $=\frac{115.0\ g}{156.6048\ g}\times100\approx73.40%$.

Answer:

73.40%