For the reaction
4 P4 + 5 S8 → 4 P4S10 ,
what is the maximum amount of P4S10
(444.545 g/mol) which could be formed from
17.06 mol (123.895 g/mol) of P4 and 7.98 mol
of S8 (256.52 g/mol)?
Answer in units of mol.
CHEMISTRY QUESTION?
2018-01-17 10:52 am
回答 (1)
2018-01-17 11:01 am
Initial number of moles of P₄ = 17.06 mol
Initial number of moles of S₈ = 7.98 mol
4 P₄ + 5 S₈ → 4 P₄S₁₀
Mole ratio P₄ : S₈ : P₄S₁₀ = 4 : 5 : 4
When 7.98 mol of S₈ completely reacts, P₄ needed = (7.98 mol) × (4/5) = 6.38 mol < 17.06 mol
Hence, P₄ is in excess, and thus S₈ is the limiting reactant/reagent which completely reacts.
Number of moles of S₈ reacted = 7.98 mol
Number of moles of P₄S₁₀ produced = (7.98 mol) × (4/5) = 6.38 mol (to 3 sig. fig.)
Initial number of moles of S₈ = 7.98 mol
4 P₄ + 5 S₈ → 4 P₄S₁₀
Mole ratio P₄ : S₈ : P₄S₁₀ = 4 : 5 : 4
When 7.98 mol of S₈ completely reacts, P₄ needed = (7.98 mol) × (4/5) = 6.38 mol < 17.06 mol
Hence, P₄ is in excess, and thus S₈ is the limiting reactant/reagent which completely reacts.
Number of moles of S₈ reacted = 7.98 mol
Number of moles of P₄S₁₀ produced = (7.98 mol) × (4/5) = 6.38 mol (to 3 sig. fig.)
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