Which substance is the limiting reactant when 16.0 g of sulfur reacts with 10.0 g of oxygen and 12.0 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)Na₂SO₄(aq)

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To determine the limiting reactant, we first need to convert the masses of the reactants to moles using their molar masses. The molar mass of sulfur (S) is approximately 32.07 g/mol, oxygen (O2) is approximately 32.00 g/mol, and sodium hydroxide (NaOH) is approximately 40.00 g/mol.- For sulfur: $16.0 \, \text{g} \, S \times \frac{1 \, \text{mol} \, S}{32.07 \, \text{g} \, S} = 0.499 \, \text{mol} \, S$- For oxygen: $10.0 \, \text{g} \, O_2 \times \frac{1 \, \text{mol} \, O_2}{32.00 \, \text{g} \, O_2} = 0.3125 \, \text{mol} \, O_2$- For sodium hydroxide: $12.0 \, \text{g} \, NaOH \times \frac{1 \, \text{mol} \, NaOH}{40.00 \, \text{g} \, NaOH} = 0.300 \, \text{mol} \, NaOH$The chemical equation shows the stoichiometry of the reaction: 2 moles of S react with 3 moles of O2 and 4 moles of NaOH. To find the limiting reactant, we compare the mole ratio of the reactants to the coefficients in the balanced equation:- For sulfur: $0.499 \, \text{mol} \, S / 2 = 0.2495$- For oxygen: $0.3125 \, \text{mol} \, O_2 / 3 = 0.1042$- For sodium hydroxide: $0.300 \, \text{mol} \, NaOH / 4 = 0.075$The smallest ratio corresponds to sodium hydroxide, indicating it is the limiting reactant because it will be consumed first according to the stoichiometry of the reaction.

Which Substance Is the Limiting Reactant When 16