Which substance is the limiting reactant when 2 g of s reacts with 3 g of o2 and 4 g of naoh 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.

S is the limiting reagent.

Explanation:

To find the limiting reactant we must first write the balanced chemical reaction. It must be correctly balanced so that we can find the proper mole ratios.

2 S (s) + 3 O2 (g) + 4 NaOH (aq) → 2 Na2SO4 (aq) + 2 H2O (l)

After this we will convert our measurements to moles. For mass we do this by dividing by the molar mass.

2g ÷ 32.06 = 0.06238mol S

3g ÷ 32.00 = 0.09375mol O₂

Now that we have the moles of each of the reactants, we can multiply them by their mole ratio with a reactant.

0.06238mol S × 2/2 = 0.06238mol H2O

0.09375mol O₂ × 2/3 = 0.06250mol H2O

S is our limiting reagent because it makes the smaller amount of moles.

it's all four options.