Which substance is the limiting reactant when 8.0 g of sulfur reacts with 12 g of oxygen and 16 g of sodium hydroxide according to the following chemical equation? 2 S(s)+ 3 O₂(g)+ 4 NaOH(aq)→ 2 Na₂SO₄(aq)+ 2 H₂O(l) A)S(s) B)O₂(g) C)NaOH(aq) D)None of these substances is the limiting reactant.

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Quizplus · Accepted Answer

To determine the limiting reactant, we first need to calculate the moles of each reactant. The molar masses are approximately 32 g/mol for sulfur (S), 32 g/mol for oxygen (O2), and 40 g/mol for sodium hydroxide (NaOH). Thus, we have:- 8.0 g S / 32 g/mol = 0.25 mol S- 12 g O2 / 32 g/mol = 0.375 mol O2- 16 g NaOH / 40 g/mol = 0.4 mol NaOHAccording to the stoichiometry of the given reaction, 2 moles of S react with 3 moles of O2 and 4 moles of NaOH. Therefore, for 0.25 moles of S, we would need 0.375 moles of O2 and 0.5 moles of NaOH. Since we only have 0.4 moles of NaOH, which is less than the required 0.5 moles, NaOH is the limiting reactant.

Which Substance Is the Limiting Reactant When 8