Answer:
a) F₅I + 3 H₂O ⇒ HIO₃ + 5 HF
b) HIO₃ 4.40×10⁻² M
HF 2.20x10⁻¹ M
Step-by-step explanation:
Once we have balanced the equation we use stoichiometry to calculate the amount of product formed. Looking at the equation, we can see that the reaction of 1 mole of H₅I with 3 moles of water yields 1 mole of HIO₃ and 5 moles of HF, respectively.
Therefore, we can calculate:
For HIO₃:
1 mole F₅I ------------------------ 1 mole HIO₃
3.91×10⁻² moles F₅I ----------- x= 3.91×10⁻² moles HIO₃
For HF:
1 mole F₅I --------------------------- 5 moles HF
3.91×10⁻² moles F₅I --------------- x= 1.95x10⁻¹ moles HF
However, these moles are contained in 888 mL, to calculate the concentration in molarity we need to calculate the moles present in 1 liter, or 1000 mL of volume (this comes from the definition of molarity).
Then, for HIO₃:
888 mL -------------------- 3.91×10⁻² moles HIO₃
1000 mL ------------------- x= 4.40×10⁻² moles HIO₃
For HF:
888 mL ------------------- 1.95x10⁻¹ moles HF
1000 mL ------------------ x= 2.20x10⁻¹ moles HF