Question
Question: What is the maximum yield of iodine that can be obtained when 1 mole of \[N{{a}_{2}}{{S}_{2}}{{O}_{8...
What is the maximum yield of iodine that can be obtained when 1 mole of Na2S2O8 reacts completely with excess iodide ion according to the equation above?
A.1 mole
B.2 moles
C.4 mole
D.6 mole
E.8 mole
Solution
The mole is defined as the method for expressing the amount of the substance. The 1 mole is defined as the amount of the substance which tends to contain 6.023×1023 elementary entities. 1 mole is equal to Avogadro number. The elementary identities means the atoms or molecules or ions.
Complete step-by-step answer: The chemical equation for the reaction is :
S2O82−+I−→SO42−+I2
In the above equation we see that the atoms are not balanced. There are 2 sulphur atoms, 8 oxygen atoms and 1 iodine atom in reactants whereas in the products side we have 1 sulphur atom, 4 oxygen atoms and 2 iodine atoms. So to balance the equation we need to multiply the iodide ion in the reactant side with 2 and the sulphate ion by 2 in the product side. So now the balanced equation is:
S2O82−+2I−→2SO42−+I2
From the balanced equation we analyse that we need 1 mole of S2O82− to give 1 mole of I2. But we need two moles of iodide ion to give 1 mole of iodine.
So according to the question the maximum yield of the iodine which can be obtained when 1 mole of Na2S2O8 reacts completely with the excess iodide ion according to the balanced equation is 1 mole.
So the correct answer is one mole that is option (A).
Note: If we need to calculate the maximum yield of iodine when 1 mole of iodide react then we need to divide 1 by 2 because from the equation we see that 2 moles of iodide ion gives 1 mole of iodine so half of iodide would give 1 mole of iodine. But if the yield of sulphate is asked from 1 mole of iodide then 1 mole would be required of iodide ion to give one mole of sulphate ion. So in these questions we need to have balanced equations to calculate the yields.