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Zigmanuir [339]
1 year ago
15

Solid iron (III) oxide reacts with hydrogen gas to form iron and water. How many grams of iron are produced when 440.23 grams of

iron(III) oxide are reacted?
Chemistry
1 answer:
Nookie1986 [14]1 year ago
4 0

When 440.23 grams of iron(III) oxide are reacted with hydrogen gas, the amount of iron produced will be 307.66 grams

<h3>Stoichiometric calculation</h3>

From the equation of the reaction:

Fe_2O_3 + 3H_2 --- > 2Fe + 3H_2O

The mole ratio of iron(III) oxide to produced iron is 1:2.

Mole of 440.23 iron(III) oxide = 440.23/159.69 = 2.76 moles

Equivalent mole of produced iron = 2.76 x 2 = 5.52 moles

Mass of 5.52 moles of iron = 5.52 x 55.8 = 307.66 grams

More on stoichiometric calculations can be found here; brainly.com/question/27287858

#SPJ1

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Answer:

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Explanation:

Hello!

In this case, since the definition of entropy in a random mixture is:

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For this silver-gold mixture we write:

\Delta S=-(n_{Au}+n_{Ag})R\Sigma[\frac{n_{Au}}{n_{Au}+n_{Ag}} *ln(\frac{n_{Au}}{n_{Au}+n_{Ag}} )+\frac{n_{Ag}}{n_{Au}+n_{Ag}} *ln(\frac{n_{Ag}}{n_{Au}+n_{Ag}} )]

By knowing the moles of gold:

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It is possible to write the aforementioned formula in terms of the variable x representing the moles of silver:

20\frac{J}{mol}=-(0.508+x)8.314\frac{J}{mol*K} \Sigma[\frac{0.508}{0.508+x} *ln(\frac{0.508}{0.508+x} )+\frac{x}{0.508+x} *ln(\frac{x}{0.508+x} )]

Which can be solved via Newton-Raphson or a solver software, in this case, I will provide you the answer:

x=n_{Ag}=21.0molAg

So the mass is:

m_{Ag}=21.0mol*\frac{107.9g}{1mol}\\ \\m_{Ag}=2,265.9g

Best regards!

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