For the reaction 2MnO_2 + 4KOH + O_2 + Cl_2 -> 2KMnO_4 + 2KCl + 2H_2O, there are 100.0 g of each reactant available. Which reactant is the limiting reagent?

Jun 20, 2016

Surely it is the base?

Explanation:

Referring directly to your balanced chemical equation, there are
$\frac{100.0 \cdot g}{86.94 \cdot g \cdot m o {l}^{-} 1} M n {O}_{2} = 1.15 \cdot m o l$

$\frac{100.0 \cdot g}{56.11 \cdot g \cdot m o {l}^{-} 1} K O H = 1.78 \cdot m o l$

$\frac{100.0 \cdot g}{32.00 \cdot g \cdot m o {l}^{-} 1} {O}_{2} = 3.13 \cdot m o l$
And,
$\frac{100.0 \cdot g}{70.9 \cdot g \cdot m o {l}^{-} 1} C {l}_{2} = 1.41 \cdot m o l$

I think the identity of the limiting reagent is fairly clear.