Answer:
1.97 moles of Na contains greater number of atoms than 6.8 × 10²² atoms of Li.
Explanation:
To solve this problem we will first calculate the number of atoms contained by 1.97 moles of Na and then will compare it with 6.8 × 10²² atoms of Li.
As we know that 1 mole of any substance contains exactly 6.022 × 10²³ particles which is also called as Avogadro's Number. So in order to calculate the number of atoms contained by 1.97 moles of Na, we will use following relation,
Moles = Number of Atoms ÷ 6.022 × 10²³ atoms.mol⁻¹
Solving for Number of Atoms,
Number of Atoms = Moles × 6.022 × 10²³ Atoms.mol⁻¹
Putting values,
Number of Atoms = 1.07 mol × 6.022 × 10²³ Atoms.mol⁻¹
Number of Atoms = 1.18 × 10²⁴ Atoms of Na
Hence,
1.97 moles of Na contains greater number of atoms than 6.8 × 10²² atoms of Li.
For this problem, we use the Hess' Law.
ΔHrxn = ∑(ν*Hf of products) - ∑(ν*Hf of reactants)
The ν represents the corresponding stoichiometric coefficients of the substances, while Hf is the heat of formation. For pure elements, Hf = 0.
Hf of Al₂O₃ = <span>−1676.4 kJ/mol
</span>Hf of Fe₂O₃ = <span>-826.0 kJ/mol
Thus,
</span>ΔHrxn = 1*−1676.4 kJ/mol + 1*-826.0 kJ/mol
<em>ΔHrxn = -2502.4 kJ/mol</em>
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