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Sagot :
Answer
The limiting reactant is iron(III) oxide (Fe2SO3)
The reactant in excess is the solid sodium (Na)
Explanation
Given:
Mass of Na = 100.0 g
Mass of Fe2O3 = 100.0 g
Equation: 6Na(s) + Fe2O3(s) ⇨ 3Na2O(s) + 2Fe(s)
What to find:
The limiting reactant and the reactant in excess.
Step-by-step solution:
Step 1: Convert the mass of the reactants given to moles using the mole formula:
[tex]Moles=\frac{Mass}{Molar\text{ }mass}[/tex]Using the atomic mass of elements in the periodic table;
The molar mass of Na = 23.0 g/mol
The molar mass of Fe2O3 = 159.69 g/mol
So,
[tex]\begin{gathered} Moles\text{ }of\text{ }Na=\frac{100.0\text{ }g}{23.0\text{ }g\text{/}mol}=4.35\text{ }mol \\ \\ Moles\text{ }of\text{ }Fe_2SO_3=\frac{100.0\text{ }g}{159.69\text{ }g\text{/}mol}=0.626\text{ }mol \end{gathered}[/tex]Step 2: Determine the limiting reactant and reactant in excess by comparing the moles in step 1 with the mole ratio of reactant in the given equation.
From the balanced equation; 1 mol Fe2SO3 = 6 mol Na
So, 0.626 mol Fe2SO4 = x mol Na
[tex]\begin{gathered} x=\frac{0.626\text{ }mol\text{ }Fe_2SO_4}{1\text{ }mol\text{ }Fe_2SO_4}\times6\text{ }mol\text{ }Na \\ \\ x=3.756\text{ }mol\text{ }Na \end{gathered}[/tex]This means 0.626 mol of Fe2SO3 will consume 3.756 mol Na out of the available 4.35 mol of Na.
Therefore, iron(III) oxide is the limiting reactant and the solid sodium is the reactant in excess.
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