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3 years ago

How many iron atoms would be produced by 4.5g of Fe2O3 reacting eith an excess of magnesium

Chemistry

2 answers:

tatiyna3 years ago

6 0

Hope this helps you.

ankoles [38]3 years ago

5 0

Answer : $0.34\times 10^{23}atoms$

Explanation : Given,

Mass of $Fe_2O_3$ = 4.5 g

Molar mass of $Fe_2O_3$ = 159.69 g/mole

Molar mass of Fe = 56 g/mole

First we have to calculate the moles of $Fe_2O_3$ .

$\text{Moles of }Fe_2O_3=\frac{\text{Mass of }Fe_2O_3}{\text{Molar mass of }Fe_2O_3}=\frac{4.5g}{159.69g/mole}=0.028moles$

Now we have to calculate the moles of Fe.

The balanced chemical reaction will be,

$Fe_2O_3+3Mg\rightarrow 2Fe+3MgO$

From the balanced reaction we conclude that

As, 1 moles of $Fe_2O_3$ react to give 2 moles of $Fe$

So, 0.028 moles of $Fe_2O_3$ react with $\frac{2}{1}\times 0.028=0.056$ moles of $Fe$

Now we have to calculate the mass of Fe.

$\text{Atoms of}Fe=\text{Moles of }Fe\times \text{Avogadro's Number}$

$\text{Atoms of}Fe=(0.056mole)\times (6.023\times 10^{23})=0.34\times 10^{23}atoms$

Therefore, the number of atoms will be $0.34\times 10^{23}atoms$

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3 years ago

What is the difference between ionization energy and ionization potential

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The octet rule states that atoms will gain, lose, or share valence electrons to get a full outer shell (orbital). Explain in you

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3 years ago

Read 2 more answers

A 100.0 mL solution containing 0.864 g of maleic acid (MW=116.072 g/mol) is titrated with 0.276 M KOH. Calculate the pH of the s

Lilit [14]

Answer:

pH = 1.32

Explanation:

H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺

This problem involves a weak diprotic acid which we can solve by realizing they amount to buffer solutions. In the first deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:

So first calculate the moles reacted and produced:

n H₂M = 0.864 g/mol x 1 mol/ 116.072 g = 0.074 mol H₂M

54 mL x 1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH

it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.

moles H₂M left = 0.074 - 0.015 = 0.059

moles HM⁻ produced = 0.015

Using the Henderson - Hasselbach equation to solve for pH:

ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325

Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.

For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.

3 0

3 years ago

Using the standard reduction potentials, Pb 2+(aq) + 2e– => Pb(s), E° = –0.13 V Fe 2+(aq) + 2e– => Fe(s), E° = –0.44 V Zn

mojhsa [17]

Answer:

Pb(s), Fe(s) and Zn(s) will reduce $Mn^{3+}$ to $Mn^{2+}$

Fe(s) and Zn(s) will reduce $Cr^{3+}$ to $Cr$

Explanation:

Standard reduction potential denotes ability to consume electrons from another species.

Hence, higher the standard reduction potential, higher will be the ability to oxidize another species.

Metal with $E_{red}^{0}$ value lower than 1.51 V will donate electron to $Mn^{3+}$ and thus reduces $Mn^{3+}$ to $Mn^{2+}$ .

So, Pb(s), Fe(s) and Zn(s) will reduce $Mn^{3+}$ to $Mn^{2+}$ .

Metal with $E_{red}^{0}$ value lower than -0.40 V will donate electron to $Cr^{3+}$ and thus reduces $Cr^{3+}$ to $Cr$ .

So, Fe(s) and Zn(s) will reduce $Cr^{3+}$ to $Cr$ .

6 0

3 years ago