Answer:
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.
