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Answer:
Fe₂O₃ is the limiting reactant.
7.57 g of MgO are formed.
Explanation:
- 3 Mg + 1 Fe₂O₃ → 2 Fe + 3 MgO
First we <u>convert the given masses of both reactants into moles</u>, using their <em>respective molar masses</em>:
- 15.6 g Mg ÷ 24.305 g/mol = 0.642 mol Mg
- 10.0 g Fe₂O₃ ÷ 159.69 g/mol = 0.0626 mol Fe₂O₃
0.0626 moles of Fe₂O₃ would react completely with (3 * 0.0626 ) 0.188 moles of Mg. As there are more Mg moles than required, Mg is the reactant in excess; thus, <em>Fe₂O₃ is the limiting reactant</em>.
We now <u>calculate how many MgO moles are produced</u>, using the <em>number of moles of the limiting reactant</em>:
- 0.0626 mol Fe₂O₃ *
= 0.188 mol MgO
Finally we <u>convert moles of MgO into grams</u>:
- 0.188 mol MgO * 40.3 g/mol = 7.57 g