Very confusing chemistry question please help!!! I'm begging you.?

[匿名]

2017-11-05 1:55 pm

How many milliliters of ozone gas at at 25.0°C and 1.00 atm pressure are needed to react with 45.00 mL of a 0.100 M aqueous solution of KI according to the chemical equation shown below?
O3(g) + 2 I-(aq) + H2O(l) → O2(g) + I2(s) + 2 OH-(aq)

This is the answer; 55.0 mL
But I don't understand how to get to it I just converted Kl to moles but how does the info about Kl help me here???

更新1:

I don't see Kl in the balanced equation though how do you know it's 1/2 O3

回答 (1)

不用客氣

2017-11-05 2:03 pm

KI(aq) → K⁺(aq) + I⁻(aq)
1 mol of KI completely dissociates in aqueous solution to give 1 mole of I⁻.
Hence, No. of moles of KI dissolved = No. of moles of I⁻ in the solution
and also, concentration of KI in the solution = Concentration of I⁻ in the solution

No. of moles of I⁻ reacted = (0.100 mol/L) × (45.00/1000 mL) = 0.0045 mol

O₃(g) + 2 I⁻(aq) + H₂O(l) → O₂(g) + I₂(s) + 2 OH⁻(aq)
According to the coefficients of the equation, 2 moles of I⁻ react with 1 mole of O₃, i.e.
1 mole of I⁻ reacts with 1/2 mole of O₃.
No. of moles of O₃ reacted with 0.0045 mole of I⁻ = (0.0045 mol) × (1/2) = 0.00225 mol

Consider the O₃ gas :
Gas law : PV = nRT
Then, V = nRT/P

Volume of O₃ gas needed, V = 0.00225 × 0.08206 × (273 + 25.0) / 1.00 L = 0.0550 L = 55.0 mL

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