Answer
$KOH$ is the limiting reactant.
Work Step by Step
1. For 1 mol of that reaction to occur, calculate the necessary mass:
- Multiply the molar mass and the balance coefficient:
$KOH$: 39.1* 1 + 16* 1 + 1.008* 1 = 56.108g/mol * 4 mol= 224.432g
$MnO_2$: 54.94* 1 + 16* 2 = 86.94g/mol * 2 mol = 173.88g
$O_2$: 16* 2 = 32g/mol * 1 mol = 32g
$Cl_2$: 35.45 * 2 = 70.90g/mol * 1 mol = 70.90g
$KOH$ needs a greater mass, so, it is the limiting reactant.
