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The Swedish chemist Karl Wilhellm was the first to produce chlorine in the lab
2NaCl + 2H2SO4 + MnO2 -----> Na2SO4 + MnSO4 + H2O + Cl2
If Dr. Wilhellm started with 50.0 g of each reactant, which reactant is the limiting reactant?

Sagot :

jcherry99

Answer:

Explanation:

Remark

Interesting que8stion. You have to figure out how many mols are present in each reactant. Since all periodic tables are different, I'm going to use rounded numbers. If it is too close, I will go further.

NaCl

Na = 23

Cl = 35.5

1 mol = 58.5 grams

given = 50.0 grams

Mols for the reaction = 50/58.5 = 0.855

H2SO4

H2       =    2*1        2

S          =   1 * 32   32

O4       =   4*16     64

1 mol  =               98   grams

mols  present = 50/98 = 0.510

MnO2

Mn     = 1 * 55    = 55

O2   =   2*16      = 32

1 mol =                  87 grams

mols available    = 50/87 = 0.5747

Discussion

Na Cl and H2SO4 both require 2 moles for every mol of Cl2 produces.

H2SO4  has 0.51 mols available for a reaction

NaCl has  0.855 moles available for a reaction

MnO2 has 0.575 moles available for a reaction.

Given those numbers 0.510 mols of H2SO4 will only produce 0.255 mols of chlorine and the rest will be reduced in a similar manner. H2SO4 is the limiting reagent (reactant).

In other words only 0.510 moles of NaCl will be used and 0.855 - 0.510 moles will be left over on the reactants side.

only 0.575 moles of MnO2 will be used and 0.065 moles will be left over.

The oddity in the result shows up because the balance numbers in the equation give a ratio of 2 to 1 for H2SO4 and NaCl The 2 belongs to the reactants and the 1 for the chlorine.

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