8. Ammonia reacts with oxygen to produce nitrogen monoxide ar water according to the following chemical reaction. 4NH_(3)(g)+5O_(2)(g)Longrightarrow 4NO(g)+6H_(2)O(l) If 128 g O_(2) reacts with excess NH3 what mass of NO is produced? 3.2 g 96g 120 g 45g

To solve this problem, we need to use the balanced chemical equation and the given information to determine the mass of NO produced.<br /><br />Given information:<br />- The balanced chemical equation is: $4NH_{3}(g)+5O_{2}(g)\Longrightarrow 4NO(g)+6H_{2}O(l)$<br />- 128 g of $O_{2}$ reacts with excess $NH_{3}$.<br /><br />Step 1: Calculate the molar mass of $O_{2}$.<br />Molar mass of $O_{2}$ = 32 g/mol<br /><br />Step 2: Calculate the number of moles of $O_{2}$.<br />Number of moles of $O_{2}$ = 128 g / 32 g/mol = 4 mol<br /><br />Step 3: Use the balanced chemical equation to determine the number of moles of NO produced.<br />According to the balanced equation, 5 moles of $O_{2}$ produce 4 moles of NO.<br />Number of moles of NO produced = (4 mol $O_{2}$) × (4 mol NO / 5 mol $O_{2}$) = 3.2 mol NO<br /><br />Step 4: Calculate the mass of NO produced.<br />Molar mass of NO = 30 g/mol<br />Mass of NO produced = 3.2 mol × 30 g/mol = 96 g<br /><br />Therefore, the correct answer is 96 g.