Multiple Choice 10 points 4Fe+3O_(2)... 2Fe_(2)O_(3) How many grams of Fe_(2)O_(3) can be produced when 1.84 moles of O_(2) reacts with excess iron (Fe)? 3928 6.12g 39.3g 1968

This is a stoichiometry problem. The balanced chemical equation is given as $4Fe+3O_{2}\cdots 2Fe_{2}O_{3}$. From this equation, we can see that 3 moles of $O_{2}$ react with 4 moles of $Fe$ to produce 2 moles of $Fe_{2}O_{3}$. Therefore, the molar ratio of $O_{2}$ to $Fe_{2}O_{3}$ is 3:2. Given that 1.84 moles of $O_{2}$ are reacting, we can calculate the moles of $Fe_{2}O_{3}$ produced by multiplying 1.84 by 2/3, which gives us 1.2267 moles of $Fe_{2}O_{3}$. The molar mass of $Fe_{2}O_{3}$ is 159.69 g/mol, so the mass of $Fe_{2}O_{3}$ produced is 1.2267 moles * 159.69 g/mol = 196.8 g. Therefore, the answer is 1968g.