Q2. Iron (III) bromide and sodium phosphate react to form a precipitate. Determine the mass of sodium phosphate required to prepare a 17.1 g precipitate
FeBr3+ Na3PO4---> NaBr+FePO4

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Ответ:

emmaphillips2

23.02.2021

Chemistry

Rate of appearance of O2 = 1.75*10^-5 M/s

Explanation:

NO2 decomposes to O2 as:

2NO2 → 2NO + O2

$Rate of reaction = -\frac{1}{2} \frac{\Delta [NO_2]}{\Delta t} =\frac{1}{2} \frac{\Delta [NO]}{\Delta t}=\frac{1}{1} \frac{\Delta [O_2]}{\Delta t}$

minus sign signifies that NO2 disappear in the reaction.

Where,

$\frac{\Delta [NO_2]}{\Delta t} = Rate of disappearance of NO_2$

$\frac{\Delta [NO]}{\Delta t} = Rate of appearance of NO$

$\frac{\Delta [O_2]}{\Delta t} = Rate of appearance of O_2$

Initial concentration of NO2 = 0.0100 M

After 100 s concentration of NO2 = 0.00650 M

Change in concentration of NO2 = (0.0100 - 0.00650)M

= 0.0035 M

$\frac{1}{2} \frac{\Delta [NO_2]}{\Delta t} =\frac{1}{1} \frac{\Delta [O_2]}{\Delta t}$

$\frac{\Delta [O_2]}{\Delta t} = \frac{1}{2} \frac{0.0035}{100}$

$\frac{\Delta [O_2]}{\Delta t} = 0.0000175 or 1.75 \times 10^{-5} M/s$

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Q2. Iron (III) bromide and sodium phosphate react to form a precipitate. Determine the mass of sodium phosphate required to prepare a 17.1 g precipitate FeBr3+ Na3PO4---> NaBr+FePO4