25.0g the Fe2O3. 2(55.8)g+3(16.0)g=159.9g. 2(55.8)g/159.6g•100%= 69.9%. 69.9%/100%•25.0g=17.5g of iron recovered

The mass of steel recovered is 17.5 grams

Explanation:

We are offered a chemical compound having chemical formula the

We know that:

Molar fixed of stole (III) oxide = 159.7 g/mol

Mass of iron = 55.8 g/mol

We room given:

Mass of stole (III) oxide = 25 g

To calculation the mass of iron recovered in given amount of steel (III) oxide, we usage unitary method:

From 159.7 grams of iron (III) oxide, fixed of stole recovered is

So, from 25 grams of steel (III) oxide, fixed of iron recovered will certainly be

Hence, the fixed of stole recovered is 17.5 grams

in the provided question, the partial press is calculation by utilizing the right gas equation, that is, pv = nrt

here, ns is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas continuous = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm

6.4 cm (max) - 25 centimeter (min)

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