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3 years ago

Calculemos la formula empirica y molecular de carbonato de hierro Fe2(co3)3.Fe.38.3%C.12.3%.0.4 92% masa 292g

Chemistry

1 answer:

DENIUS [597]3 years ago

7 0

Respuesta:

292 g / mol;

Por favor, consulte la explicación.

Explicación:

El número de átomos de cada elemento en el compuesto: Fe2 (co3) 3.

Fe = 2; C = 3; 0 = 3 * 3 = 9

Fe2 = 112 g / mol

C = 12 g / mol

O = 16 g /

Masa molecular = ((112 + (12 * 3) + (16 * 9)

= (112 + 36 + 144)

= 292 g / mol.

Total = 2 + 3 + 9 = 14

Fe2 = 112/292 * 100% = 38,356%

C = 36/292 * 100% = 12,328 * '

O = (16 * 9) / 292 * 100% = 49,32%

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3 years ago

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Calculate the mass of Na2O that can be produced by the chemical reaction of 4.0 grams of sodium with excess oxygen in the reacti

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Answer:

The mass of Na₂O that can be produced by the chemical reaction of 4.0 grams of sodium with excess oxygen in the reaction is 5.39 grams.

Explanation:

You know the balanced reaction:

4 NA + O₂ ⟶ 2 Na₂O

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction) react and are produced:

Na: 4 moles
O₂: 1 mole
Na₂O: 2 moles

Being:

Na: 23 g/ole
O: 16 g/mole

the molar mass of the compounds participating in the reaction is:

Na: 23 g/mole
O₂: 2*16 g/mole= 32 g/mole
Na₂O: 2*23 g/mole +16 g/mole= 62 g/mole

Then by stoichiometry of the reaction they react and are produced:

Na: 4 moles* 23 g/mole= 92 g
O₂: 1 mole*32 g/mole= 32 g
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Then you can apply the following rule of three: if 92 grams of Na produce 124 grams of Na₂O, 4 grams of Na, how much mass of Na₂O does it produce?

$massofNa_{2}O =\frac{4 grams of Na*124 gramsofNa_{2}O }{92 grams of Na}$

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2 years ago

Copper(II) sulfide is formed when copper and sulfur are heated together. In this reaction,

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Answer:

The mass of copper(II) sulfide formed is:

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Explanation:

The Balanced chemical equation for this reaction is :

$Cu(s) + S\rightarrow CuS$

given mass= 54 g

Molar mass of Cu = 63.55 g/mol

$Moles = \frac{given\ mass}{Molar\ mass}$

$moles=\frac{54}{63.55}$

Moles of Cu = 0.8497 mol

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Molar mass of S = 32.06 g/mol

$Moles = \frac{given\ mass}{Molar\ mass}$

$moles=\frac{42}{32.06}$

Moles of S = 1.31 mol

Limiting Reagent : The reagent which is present in less amount and consumed in a reaction

First find the limiting reagent :

$Cu + S\rightarrow CuS$

1 mol of Cu require = 1 mol of S

0.8497 mol of Cu should require = 1 x 0.8497 mol

= 0.8497 mol of S

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S Required = 0.8497 mol

S is present in excess and Cu is limiting reagent

All Cu is consumed in the reaction

Amount Cu will decide the amount of CuS formed

$Cu + S\rightarrow CuS$

1 mole of Cu gives = 1 mole of Copper sulfide

0.8497 mol of Cu = 1 x 0.8497 mole of Copper sulfide

= 0.8497

Molar mass of CuS = 95.611 g/mol

$Moles = \frac{given\ mass}{Molar\ mass}$

$0.8497 = \frac{given\ mass}{95.611}$

Mass of CuS = 0.8497 x 95.611

= 81.24 g

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Answer:

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