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.

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. completely dissociates to and ions. a is incorrect because does not fully ionize to and ions and some remain as the compound . c and d are incorrect because, and do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise,   if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both and have a complete octet so can not accept any electrons which disqualifies them as acids.