Question

Balance the chemical equation given below, and determine the number of grams of MgO are needed to produce 10.0g of Fe₂0₃ :

____ MgO + ____Fe ____ Fe₂0₃​+ ____Mg

Answer 1

To balance the equation and calculate the grams of MgO needed to produce 10.0g of Fe2O3, 3.78 grams of MgO are required, based on the balanced reaction and stoichiometry calculations.

Step-by-step explanation:

To balance the chemical equation and determine the number of grams of MgO needed to produce 10.0g of Fe₂O₃, we first balance the equation for the reduction of iron(III) oxide by magnesium:

• 3 MgO + 2 Fe → Fe₂O₃ + 3 Mg

Now, let's calculate the moles of Fe₂O₃ formed:

1. Find the molar mass of Fe₂O₃ which is 159.70 g/mol.
2. Calculate moles of Fe₂O₃ produced from 10.0 g: (10.0 g Fe₂O₃) / (159.70 g Fe₂O₃/mol) = 0.0626 mol Fe₂O₃.

According to the balanced chemical equation, the mole ratio of MgO to Fe₂O₃ is 3:2. Using the mole ratio and the moles of Fe₂O₃, calculate the moles of MgO required:

0.0626 mol Fe₂O₃ x (3 mol MgO / 2 mol Fe₂O₃) = 0.0939 mol MgO

Finally, convert moles of MgO to grams:

1. The molar mass of MgO is approximately 40.30 g/mol.
2. Grams of MgO needed: (0.0939 mol MgO) x (40.30 g MgO/mol) = 3.7837 g MgO.

Thus, 3.78 grams of MgO are needed to produce 10.0 g of Fe₂O₃.