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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?

Answer 1

Wyatt Adkins

Surely it is the base?

Referring directly to your balanced chemical equation, there are #(100.0*g)/(86.94*g*mol^-1)MnO_2=1.15*mol#

#(100.0*g)/(56.11*g*mol^-1)KOH=1.78*mol#

#(100.0*g)/(32.00*g*mol^-1)O_2=3.13*mol# And, #(100.0*g)/(70.9*g*mol^-1)Cl_2=1.41*mol#

I believe it is fairly obvious who the limiting reagent is.

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Answer 2

Cameron Andrews

To find the limiting reagent, first, calculate the number of moles of each reactant using their respective molar masses. Then, use the stoichiometric coefficients from the balanced equation to determine the amount of product that can be formed from each reactant. The reactant that produces the least amount of product is the limiting reagent.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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

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?