10. Given the reaction: 4NH_(3)(g)+5O_(2)(g)arrow 4NO(g)+6H_(2)O When 8 moles of ammonia (NH_(3)) reacts, the total number of grams of NO formed is __

To solve this problem, we need to use the stoichiometry of the balanced chemical equation to determine the amount of NO formed when 8 moles of NH3 react.<br /><br />Given information:<br />- Balanced chemical equationNH_{3}(g)+5O_{2}(g)\rightarrow 4NO(g)+6H_{2}O$<br />- Amount of NH3: 8 moles<br /><br />Step 1: Determine the mole ratio between NH3 and NO.<br />From the balanced equation, we can see that 4 moles of NH3 produce 4 moles of NO.<br />Mole ratio of NH3 to NO = 4:4 = 1:1<br /><br />Step 2: Calculate the amount of NO formed.<br />Since the mole ratio of NH3 to NO is 1:1, 8 moles of NH3 will produce 8 moles of NO.<br /><br />Step 3: Convert the amount of NO from moles to grams.<br />Molar mass of NO = 14.01 g/mol (N) + 16.00 g/mol (O) = 30.01 g/mol<br />Mass of NO = 8 moles × 30.01 g/mol = 240.08 grams<br /><br />Therefore, when 8 moles of ammonia (NH3) react, the total number of grams of NO formed is 240.08 grams.