Naric acid decomposes in light to form nitrogen dioxide, water, and oxygen according to the following chemical equati 4HNO_(3)(l)xrightarrow (light)4NO_(2)(g)+2H_(2)O(l)+O_(2)(g) Naric acid If 2.6 g nitric acid decomposes, what is the total volume of oxygen gas produced? 0.091 0.221 111 0.060

To solve this, we need to calculate the volume of oxygen gas produced when 2.6 g of nitric acid decomposes.<br /><br />### Step 1: Write down the balanced chemical equation<br />The decomposition reaction is:<br />\[<br />4HNO_3(l) \xrightarrow{\text{light}} 4NO_2(g) + 2H_2O(l) + O_2(g)<br />\]<br /><br />From the equation, **4 moles of \( HNO_3 \)** produce **1 mole of \( O_2 \)**.<br /><br />---<br /><br />### Step 2: Calculate the number of moles of \( HNO_3 \)<br />The molar mass of \( HNO_3 \) is:<br />\[<br />M(HNO_3) = 1 + 14 + (16 \times 3) = 63 \, \text{g/mol}.<br />\]<br /><br />The number of moles of \( HNO_3 \) in 2.6 g is:<br />\[<br />\text{moles of } HNO_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{2.6}{63} \approx 0.04127 \, \text{mol}.<br />\]<br /><br />---<br /><br />### Step 3: Use stoichiometry to find moles of \( O_2 \)<br />From the balanced equation, 4 moles of \( HNO_3 \) produce 1 mole of \( O_2 \). Therefore:<br />\[<br />\text{moles of } O_2 = \frac{\text{moles of } HNO_3}{4} = \frac{0.04127}{4} \approx 0.01032 \, \text{mol}.<br />\]<br /><br />---<br /><br />### Step 4: Calculate the volume of \( O_2 \) at standard conditions<br />At standard temperature and pressure (STP), 1 mole of gas occupies 22.4 L. Thus:<br />\[<br />\text{Volume of } O_2 = \text{moles of } O_2 \times 22.4 = 0.01032 \times 22.4 \approx 0.231 \, \text{L}.<br />\]<br /><br />---<br /><br />### Final Answer:<br />The total volume of oxygen gas produced is approximately **0.221 L**.<br /><br />Correct option: **0.221**