Question

Part A

Answer 1

Answer:

KOH is limiting reactant

Explanation:

To solve this question, we need to convert the mass of each reactant to moles. Then, using the balanced reaction, we can find the limiting reactant:

Moles S -Molar mass: 32g/mol-:

14g S * (1mol / 32g) = 0.44mol

Moles O₂ -Molar mass: 32g/mol-:

24g O₂ * (1mol / 32g) = 0.75mol

Moles KOH -Molar mass: 56g/mol-:

28g KOH * (1mol / 56g) = 0.5mol

For a complete reaction of the KOH are required:

0.5mol KOH * (2mol S / 4mol KOH) = 0.25mol S

0.5mol KOH * (3mol O₂ / 4mol KOH) = 0.38mol O₂

As there are 0.44 moles of S and 0.38moles of oxygen,

KOH is limiting reactant