S₂O₈²⁻
(aq) + 2I⁻
(aq) → I₂(aq) + 2SO₄
²⁻(aq)
2S₂O₃²⁻
(aq) + I₂(aq) → S₄O₆²⁻
(aq) + 2I⁻
(aq)
<u>Explanation:</u>
S₂O₈²⁻
(aq) + 2I⁻
(aq) → I₂(aq) + 2SO₄
²⁻(aq)
To measure the rate of this reaction we must measure the rate of concentration change of one of the reactants or products. To do this, we will include (to the reacting S₂O₈
²⁻ and I⁻
i) a small amount of sodium thiosulfate, Na₂S₂O₃,
ii) some starch indicator.
The added Na₂S₂O₃ does not interfere with the rate of above reaction, but it does consume the I₂ as soon as it is formed.
2S₂O₃²⁻
(aq) + I₂(aq) → S₄O₆²⁻
(aq) + 2I⁻
(aq)
This reaction is much faster than the previous, so the conversion of I2 back to I⁻ is essentially instantaneous.
![rate = \frac{dI2}{dt} = \frac{1/2 [S2O3^2^-]}{t}](https://tex.z-dn.net/?f=rate%20%3D%20%5Cfrac%7BdI2%7D%7Bdt%7D%20%3D%20%5Cfrac%7B1%2F2%20%5BS2O3%5E2%5E-%5D%7D%7Bt%7D)
Answer:
15.35 g of (NH₄)₃PO₄
Explanation:
First we need to look at the chemical reaction:
3 NH₃ + H₃PO₄ → (NH₄)₃PO₄
Now we calculate the number of moles of ammonia (NH₃):
number of moles = mass / molecular wight
number of moles = 5.24 / 17 = 0.308 moles of NH₃
Now from the chemical reaction we devise the following reasoning:
if 3 moles of NH₃ are produce 1 mole of (NH₄)₃PO₄
then 0.308 moles of NH₃ are produce X moles of (NH₄)₃PO₄
X = (0.308 × 1) / 3 = 0.103 moles of (NH₄)₃PO₄
mass = number of moles × molecular wight
mass = 0.103 × 149 = 15.35 g of (NH₄)₃PO₄
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