In the reaction 2 fe + 3 s → fe2s3 how much iron is required to react with 48.0 g of sulfur? 1. 24.8 g 2. 55.8 g 3. 83.8 g 4. 111 g 5. 126 g

2. 55.8 g
    The balanced equation indicates that for every 3 moles of Sulfur, 2 moles of Iron is required. So let's start by looking up their respective atomic weights:
 Atomic weight sulfur = 32.065
 Atomic weight iron = 55.845

   Moles sulfur = 48.0 g / 32.065 g/mol = 1.496959301 mol
 Moles iron needed = (1.496959301 / 3) * 2 = 0.997972868
 Grams iron = 0.997972868 mol * 55.845 g/mol = 55.73179479 g.

   Of the available choices, option "2" with 55.8 grams comes the closest. The difference between that value and my calculated value can be attributed to the point in time where rounding occurs. For instance, the number of moles of sulfur could have been easily rounded to 1.50 moles, in which case, 1 mole of iron would have been calculated which comes to 55.8 grams when rounded to 1 decimal place.