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Consider the following reaction
$$\ce{4H2O + 5I2O4 -> 8HIO3 + I2}$$
If 0.167 mol of $\ce{H2O}$ is added to 0.200 mol of $\ce{I2O4}$ what amount (moles) of $\ce{HIO3}$ can be formed, it the reaction goes to completion?
(A) 0.084 mol
(B) 0.125 mol
(C) 0.320 mol
(D) 0.334 mol
Can someone please explain the answer to this problem. I have been at it for ages and keep coming up with an answer close to C but not close enough to count it! Help!
To solve such questions, first find out the limiting reagent. Limiting reagent is the reactant that get over first during the reaction.
Here, it can be observed that the limiting reagent is I2O4. 0.2 moles of the compound requires only 0.16 moles of H2O. It is seen that H20 is in excess (0.07 moles is extra)
On further calculation using proportionality, we see that 0.2 moles of I2O4 gives 0.32 moles HIO3.
