Answer:
2HI + K2SO3=>2KI+H2SO3
Explanation:When aqueous hydroiodic acid and aqueous potassium sulfite are mixed the products obtained are potassium iodide and sulfurous acid.Both reactants are ionic compounds and they undergo double replacement reaction.In a double replacement reaction the parts of the ionic compounds are changed.The product is obtained by combinig cation of one compound with anion of other compound.so in above reaction sulfurous acid is obtained which is in gaseous form and potassium iodide is an ionic compound.
CaCl2 and KCl are both salts which dissociate in water
when dissolved. Assuming that the dissolution of the two salts are 100 percent,
the half reactions are:
<span>CaCl2 ---> Ca2+ + 2 Cl-</span>
KCl ---> K+ + Cl-
Therefore the total Cl- ion concentration would be coming
from both salts. First, we calculate the Cl- from each salt by using stoichiometric
ratio:
Cl- from CaCl2 = (0.2 moles CaCl2/ L) (0.25 L) (2 moles
Cl / 1 mole CaCl2)
Cl- from CaCl2 = 0.1 moles
Cl- from KCl = (0.4 moles KCl/ L) (0.25 L) (1 mole Cl / 1
mole KCl)
Cl- from KCl = 0.1 moles
Therefore the final concentration of Cl- in the solution
mixture is:
Cl- = (0.1 moles + 0.1 moles) / (0.25 L + 0.25 L)
Cl- = 0.2 moles / 0.5 moles
<span>Cl- = 0.4 moles (ANSWER)</span>
