If 25.1 grams of iron and 47.8 grams of chlorine combine to form FeCl3, how many grams of FeCl3 must form?

Answer 1

Step 1 - Write out the balanced chemical equation.

2Fe + 3Cl₂ --> 2FeCl₃

Step 2 - Determine the limiting reagent by taking the given amounts of reactants in grams and converting them to moles of product.

25.1 g of Fe x (1 mole of Fe/55.845 g Fe) x (2 moles of FeCl₃/ 2 moles of Fe) = 0.449458 moles of FeCl₃

47.8 g of Cl₂ x (1 mole of Cl₂/70.9 g Fe) x (2 moles of FeCl₃/ 3 moles of Cl₂) = 0.449459 moles of FeCl₃

Since we had to go to six decimal places to see the difference, you can use either value. However, in this case Fe is technically the limiting reagent.

Step 3 - Convert moles of FeCl₃ to grams of FeCl₃

0.449 moles FeCl₃ x (162.195 g of FeCl₃/ 1 mole of FeCl₃) = 72.825555g

Round answer to 3 significant figures since information in the problem started with 3 sig figs to get 72.8 g of FeCl₃.

Hope this helped!