Question

Which substance is the limiting reactant when 12 g of sulfur reacts with 18 g of oxygen and 24 g of sodium hydroxide according to the following chemical equation?

Answer 1

The limiting reactant in the reaction is sulfur (S).

Step-by-step explanation:

The limiting reactant is the substance that is completely consumed in a chemical reaction, thereby limiting the amount of product that can be formed. In order to determine the limiting reactant, we need to compare the amount of each reactant to the stoichiometric coefficients in the balanced chemical equation.

First, let's calculate the number of moles of each reactant:

• Sulfur (S): molar mass = 32 g/mol, 12 g S = 0.375 mol
• Oxygen (O₂): molar mass = 32 g/mol, 18 g O₂ = 0.563 mol
• Sodium hydroxide (NaOH): molar mass = 40 g/mol, 24 g NaOH = 0.6 mol

Now, we can compare the mole ratios of the reactants to the balanced chemical equation:

• S:O₂ = 0.375 mol : 0.563 mol = 1:1
• S:NaOH = 0.375 mol : 0.6 mol = 1:1.6

Based on these calculations, it is clear that the limiting reactant is sulfur (S) because it will be completely consumed before either oxygen or sodium hydroxide.

Answer 2

NaOH is the limiting reactant

Further explanation

Given

12 g of sulfur reacts with 18 g of oxygen and 24 g of sodium hydroxide

Required

The limiting reactant

Solution

Reaction

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

• mol S(MW=32.065 g/mol) :

`\tt (12)/(32.065)=0.374`

• mol O₂(MW=32 g/mol)

`\tt (18)/(32)=0.5625`

• mol NaOH(MW=40 g/mol)

`\tt (24)/(40)=0.6`

A method that can be used to find limiting reactants is to divide the number of moles of known substances by their respective coefficients, and the smallest ratio becomes a limiting reactant

S : O₂ : NaOH =

`\tt (0.374)/(2)/ (0.5625)/(3)/ (0.6)/(4)=0.187/ 0.1875/ 0.15\Rightarrow NaOH~smallest~ratio`