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3 years ago

How many grams of FeO would be needed to make 234.2 grams of Fe in the chemical reaction 4FeO = Fe3O4 + Fe? Show your work.

Chemistry

2 answers:

lidiya [134]3 years ago

6 0

Answer:

1,203.84 grams of FeO will be needed.

Explanation:

$4FeO\rightarrow Fe_3O_4 + Fe$

Moles of iron = $\frac{234.2 g}{56 g/mol}=4.18 mol$

According to reaction, 1 mol of iron is obtained from 4 moles of ferrous oxide.

Then 4.18 moles of iron will be obtained from:

$\frac{4}{1}\times 4.18 mol=16.72 mol$ of ferrous oxide

Mass of 16.72 moles of ferrous oxide:

16.72 mol × 72 g/mol = 1,203.84 g

1,203.84 grams of FeO will be needed.

Scorpion4ik [409]3 years ago

4 0

The grams of FeO that would be needed to make 234.2 grams of Fe is

1204.42 grams

calculation

4 FeO → Fe₃O₄ +Fe

Step 1: find the moles of fe

moles = mass /molar mass

from periodic table the molar mass of Fe = 56 g/mol

moles = 234.2 g/56 g/mol = 4.182 moles

Step 2: use the mole ratio to determine the moles of FeO

FeO: Fe is 4:1 therefore the moles of FeO =4.182 moles x4 =16.728 moles

Step 3: find the mass of FeO

mass = moles x molar mass

The molar mass of FeO = 56 +16 = 72 g /mol

mass = 16.728 moles x 72 g/mol= 1204.42 grams

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Answer:

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2 years ago

Which of the following is an oxidation-reduction reaction? Fe2O3 3CO Right arrow. 2Fe 3CO2 CuSO4 2NaOH Right arrow. Cu(OH)2 Na2S

Damm [24]

The reaction, Fe2O3 + 3CO------> 2Fe + 3CO2 is an oxidation-reduction reaction.

An oxidation-reduction reaction is a reaction in which there is a change in oxidation number from left to right in the reaction. This is because, a specie is oxidized and another specie is reduced.

In the reaction; Fe2O3 + 3CO------> 2Fe + 3CO2, we can see that the oxidation number of iron decreased from +3 on the left hand side to zero on the right hand side. The oxidation number of carbon was increased from + 2 to +4.

Learn more: brainly.com/question/10079361

6 0

2 years ago

i am begging anyone to help me with this! (all tutors i've asked said they can't solve it but i need someone to help me out) - i

9966 [12]

First, we need to calculate how much energy we will get from this combustion.

Assuming the combustion is complete, we have the octane reacting with O₂ to form only water and CO₂, so:

$C_8H_{18}+O_2\to CO_2+H_2O$

We need to balance the reaction. Carbon only appear on two parts, so, we can start by it:

$C_8H_{18}+O_2\to8CO_2+H_2O$

Now, we balance the hydrogen:

$C_8H_{18}+O_2\to8CO_2+9H_2O$

And in the end, the oxygen:

$C_8H_{18}+\frac{25}{2}O_2\to8CO_2+9H_2O$

We can multiply all coefficients by 2 to get integer ones:

$2C_8H_{18}+25O_2\to16CO_2+18H_2O$

Now, we need to use the enthalpies of formation to get the enthalpy of reaction of this reaction.

The enthalpy of reaction can be calculated by adding the enthalpies of formation of the products multiplied by their stoichiometric coefficients and substracting the sum of enthalpies of formation of the reactants multiplied by their stoichiometric coefficients.

For the reactants, we have (the enthalpy of formation of pure compounds is zero, which is the case for O₂):

$\begin{gathered} \Delta H\mleft\lbrace reactants\mright\rbrace=2\cdot\Delta H\mleft\lbrace C_8H_{18}\mright\rbrace+25\cdot\Delta H\mleft\lbrace O_2\mright\rbrace \\ \Delta H\lbrace reactants\rbrace=2\cdot(-250.1kJ)+25\cdot0kJ \\ \Delta H\lbrace reactants\rbrace=-500.2kJ+0kJ \\ \Delta H\lbrace reactants\rbrace=-500.2kJ \end{gathered}$

For the products, we have:

$\begin{gathered} \Delta H_{}\mleft\lbrace product\mright\rbrace=16\cdot\Delta H\lbrace CO_2\rbrace+18\cdot\Delta H\lbrace H_2O\rbrace \\ \Delta H_{}\lbrace product\rbrace=16\cdot(-393.5kJ)+18\cdot(-285.5kJ) \\ \Delta H_{}\lbrace product\rbrace=-6296kJ-5139kJ \\ \Delta H_{}\lbrace product\rbrace=-11435kJ \end{gathered}$

Now, we substract the rectants from the produtcs:

$\begin{gathered} \Delta H_r=\Delta H_{}\lbrace product\rbrace-\Delta H\lbrace reactants\rbrace \\ \Delta H_r=-11435kJ-(-500.2kJ) \\ \Delta H_r=-10934.8kJ \end{gathered}$

Now, this enthalpy of reaction is for 2 moles of C₈H₁₈, so for 1 mol of C₈H₁₈ we have half this value:

$\Delta H_c=\frac{1}{2}\Delta H_r=\frac{1}{2}\cdot(-10934.8kJ)=-5467.4kJ$

Now, we have 100 g of C₈H₁₈, and its molar weight is approximately 114.22852 g/mol, so the number of moles in 100 g of C₈H₁₈ is:

$\begin{gathered} M_{C_8H_{18}}=\frac{m_{C_8H_{18}}}{n_{C_8H_{18}}} \\ n_{C_8H_{18}}=\frac{m_{C_8H_{18}}}{M_{C_8H_{18}}}=\frac{100g}{114.22852g/mol}\approx0.875438mol \end{gathered}$

Since we have approximately 0.875438 mol, and 1 mol releases -5467.4kJ when combusted, we have:

$Q=-5467.4kJ/mol\cdot0.875438mol\approx-4786.37kJ$

Now, for the other part, we need to calculate how much heat it is necessary to melt a mass, m.

First, we have to heat the ice to 0 °C, so:

$\begin{gathered} Q_1=m\cdot2.010J/g.\degree C\cdot(0-(-10))\degree C \\ Q_1=m\cdot2.010J/g\cdot10 \\ Q_1=m\cdot20.10J/g \end{gathered}$

Then, we need to melt all this mass, so we use the latent heat now:

$Q_2=n\cdot6.03kJ/mol$

Converting mass to number of moles of water we have:

$\begin{gathered} M=\frac{m}{n} \\ n=\frac{m}{M}=\frac{m}{18.01528g/mol} \end{gathered}$

So:

$Q_2=\frac{m}{18.01528g/mol}_{}\cdot6.03kJ/mol\approx m\cdot0.334716kJ/g$

Adding them, we have a total heat of:

$\begin{gathered} Q_T=m\cdot20.10J/g+m\cdot0.334716kJ/g \\ Q_T=m\cdot0.02010kJ/g+m\cdot0.334716kJ/g \\ Q_T=m\cdot0.354816kJ/g \end{gathered}$

Since we have a heat of 4786.37 kJ form the combustion, we input that to get the mass (the negative sign is removed because it only means that the heat is released from the reaction, but now it is absorbed by the ice):

$\begin{gathered} 4786.37kJ=m\cdot0.354816kJ/g \\ m=\frac{4786.37kJ}{0.354816kJ/g}\approx13489g\approx13.5\operatorname{kg} \end{gathered}$

Since we have a total of 20kg of ice, we can clculate the percent using it:

$P=\frac{13.5\operatorname{kg}}{20\operatorname{kg}}=0.675=67.5\%$

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How many moles of MgSiO3 are in 237g of the compound?

Anna35 [415]

For this, we first calculate molecular weight of MgSiO₃:
Atomic masses:
Mg = 24
Si = 28
O = 16

Mr = 24 + 28 + 16 x 3
Mr = 100

moles = mass / Mr
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3 years ago