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

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) :

$\tt \dfrac{12}{32.065}=0.374$

• mol O₂(MW=32 g/mol)

$\tt \dfrac{18}{32}=0.5625$

• mol NaOH(MW=40 g/mol)

$\tt \dfrac{24}{40}=0.6$

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 =

$\tt \dfrac{0.374}{2}\div \dfrac{0.5625}{3}\div \dfrac{0.6}{4}=0.187\div 0.1875\div 0.15\Rightarrow NaOH~smallest~ratio$