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.
