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



Answer from: Quest

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



Answer from: Quest

6.4 cm (max) - 25 centimeter (min)


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