3 years ago

WITH BRAINLIEST

Chemistry

1 answer:

TEA [102]3 years ago

4 0

Answer:

i want go with option 2

Explanation:

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A student determines the iron(III) content of a solution by first precipitating it as iron(III) hydroxide, and then decomposing

Semmy [17]

Answer:

3.3535 g

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For iron(III) nitrate :

Molarity = 0.404 M

Volume = 52.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 52.0×10⁻³ L

Thus, moles of iron(III) nitrate :

$Moles=0.404 \times {52.0\times 10^{-3}}\ moles$

Moles of iron(III) nitrate = 0.021 moles

1 mole of iron(III) nitrate forms 1 mole of iron(III) hydroxide which further forms 1 mole of iron(III) oxide.

Thus, moles of iron(III) oxide formed from 0.021 moles of iron(III) nitrate = 0.021 moles

Also, molar mass of iron(III) oxide = 159.69 g/mol

So, mass of iron(III) oxide = 0.021 moles × 159.69 g/mol = 3.3535 g

6 0

3 years ago