2.4 g of magnesium metal reacted vigorously when heated with excess iron(III) oxide, Fe2O3, according to the following equation:3Mg(s) + Fe2O3(s) → 2Fe(s) + 3MgO(s) Determine the mass of solid iron produced in the reaction
Question
2.4 g of magnesium metal reacted vigorously when heated with excess iron(III) oxide, Fe2O3, according to the following equation:3Mg(s) + Fe2O3(s) → 2Fe(s) + 3MgO(s) Determine the mass of solid iron produced in the reaction
Solution
To solve this problem, we need to use stoichiometry, which is a method in chemistry that uses balanced chemical equations to calculate quantities of reactants or products.
Step 1: Write down the balanced chemical equation: 3Mg(s) + Fe2O3(s) → 2Fe(s) + 3MgO(s)
Step 2: Calculate the molar mass of magnesium (Mg). From the periodic table, the molar mass of Mg is approximately 24.31 g/mol.
Step 3: Calculate the number of moles of Mg in 2.4 g. Number of moles = mass / molar mass = 2.4 g / 24.31 g/mol = 0.0987 mol
Step 4: From the balanced chemical equation, we can see that 3 moles of Mg produce 2 moles of Fe. Therefore, the number of moles of Fe produced is (2/3) * 0.0987 mol = 0.0658 mol.
Step 5: Calculate the mass of Fe produced. The molar mass of Fe is approximately 55.85 g/mol. Mass = number of moles * molar mass = 0.0658 mol * 55.85 g/mol = 3.68 g
Therefore, the mass of solid iron produced in the reaction is approximately 3.68 g.
Similar Questions
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