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2 years ago

Iron reacts slowly with oxygen and water to form a compound commonly called rust (Fe₂O₃?4H₂O). For 45.2 kg of rust, calculate

1 answer:

4 0

Rust (Fe2O3. 4H2O) is formed when iron interacts slowly with oxygen and water. Mass of Fe in grams is 2.18 x 10⁴ g.

<h3>What is the explanation?</h3>

There are 2 moles of Fe atoms in 1 mole of Fe2O3-4H2O. The number of moles of Fe atoms in 45.2 kg rust is shown below.

Moles of Fe = 195.01 mol Fe₂O₃.4H₂O ( $\frac{2 mol Fe }{1 mol Fe₂O₃.4H₂O}$ )

Moles of Fe = 390.02 mol Fe

Multiply the calculated number of moles of iron, Fe, by its molar mass which is 55.85 $\frac{g}{mol}$

Mass of Fe = 390.02 mol Fe ( $\frac{55.85 g Fe}{mol Fe}$ )

Mass of Fe = 2.18 x 10⁴ g Fe

Avogadro's number (6.022 x 1023) of molecules (or formula units) make up one mole of a substance (ionic compound). The mass of 1 mole of a chemical is indicated by its molar mass. It provides you with the amount of grams per mole of a substance, to put it another way.

To learn more about moles visit:

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Determine ΔH for the reaction CaCO3 → CaO + CO2 given these data: 2 Ca + 2 C + 3 O2 → 2 CaCO3 ΔH = −2,414 kJ C + O2 → CO2 ΔH = −

kicyunya [14]

Answer:

The ΔH for the reaction is -456.5 KJ

Explanation:

Here we want to determine ΔH for the reaction;

Mathematically;

ΔH = ΔH(product) - ΔH(reactant)

In the case of the first reaction;

ΔH = ΔH(CaO) + ΔH(CO2) - ΔH(CaCO3) ...........................(*)

From the other reactions, we can get the respective ΔH for the individual molecule in the reaction

In second reaction;

Kindly note that for elements, molecule of gases, ΔH = 0

What this means is that throughout the solution;

ΔH(Ca) = 0 KJ

ΔH(O2) = 0 KJ

ΔH(C) = 0 KJ

Thus, in writing the equation for the subsequent chemical reactions, we shall need to write and equate the overall ΔH for the reaction to that of the product alone

So in the second reaction

ΔH = 2ΔH(CaCO3)

Thus;

-2414/2 = ΔH(CaCO3)

ΔH(CaCO3) = -1,207 KJ

Moving to the third reaction, we have;

ΔH = ΔH(CO2)

Hence ΔH(CO2) = -393.5 KJ

For the last reaction;

ΔH = ΔH(CaO)

Hence ΔH(CaO) = -1270 KJ

Going back to equation *

ΔH = ΔH(CaO) + ΔH(CO2) - ΔH(CaCO3)

Using the values of the ΔH of the respective molecules given above,

ΔH = -1270 + (-393.5) - (-1207)

ΔH = -456.5 KJ

8 0

3 years ago

I have 345mL of a 1.5M NaCl solution. If I boil the water until the volume of the solution is 250mL, what will the molarity of t

Karo-lina-s [1.5K]

Answer is
C. 2.07 M

For explanation

M1V1 = M2V2

M2 = (M1V1)/V2

M2 = (1.5M x 345ml) / 250 ml

:. M2 = 2.07 M

8 0

3 years ago

Which characteristic of a substance is considered a chemical property?

slavikrds [6]

B. It’s reactivity.

8 0

1 year ago

Is there a qualitative difference between the enthalpy of a phase transition versus the enthalpy of a heating or cooling process

Vesna [10]

Answer:

Explanation:

given that, enthalpy is a state function, that means it depends only on the initial and final states, there is no difference between the enthalpy of a phase transition versus the enthalpy of a heating or cooling process, when the cooling or heating process finish in a change of phase.

It does not matter which way we take to cool or heat the substances the Enthalpy of this process will be the same.

3 0

3 years ago

When 500.0 g of water is decomposed by electrolysis and the yield of hydrogen is only 75.3%, how much hydrogen chloride can be m

Evgen [1.6K]

The amount of hydrogen chloride that can be made is 1064 g

Why?

The two reactions are:

2H₂O → 2H₂ + O₂ 75.3 % yield

H₂ + Cl₂ → 2HCl 69.8% yield

We have to apply a big conversion factor to go from grams of water (The limiting reactant), to grams of HCl, the final product. We have to be very careful with the coefficients and percentage yields!

$500.0gH_2O*\frac{1moleH_2O}{18.01 gH_2O}*\frac{2 moles H_2}{2 moles H_2O}*\frac{2.015g H_2}{1 mole H_2}*\frac{75.3 actual g}{100 theoretical g}=42.12 g H_2$

$42.12H_2*\frac{1 mole H_2}{2.015gH_2}*\frac{2 moles HCl}{1 mole H_2}*\frac{36.46g}{1 mole HCl}*\frac{69.8 actualg}{100 theoreticalg} =1064gHCl$

Have a nice day!

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7 0

3 years ago