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maks197457 [2]
2 years ago
10

Mining companies use this reaction to obtain iron from iron ore: Fe2O3(s)+3CO(g) → 2Fe(s)+3CO2(g) The reaction of 177 g of Fe2O3

with 82.8 g of CO produces 72.7 g of Fe. Calculate the theoretical yield of solid iron.
ANSWER: 110. g Fe
Chemistry
1 answer:
dezoksy [38]2 years ago
7 0

Answer:

The theoretical yield of solid iron is 110.05 grams

Explanation:

<u>Step 1</u>: The balanced equation

Fe2O3(s)+3CO(g) → 2Fe(s)+3CO2(g)

This means for 1 mole of Fe2O3 consumed, there is 3 moles of CO needed to produce 2 moles of Fe and 3 moles of CO2

<u>Step 2</u>: given data

Mass of Fe2O3 = 177g

MM of Fe2O3 = 159.69g/mole

Mass of CO = 82.8g

MM of CO = 28.01

Mass of CO2 = 72.7g

MM of CO2 = 44.01 g/mole

Mass of Fe = TO BE DETERMINED

MM of Fe = 55.845 g/mole

<u>Step 3:</u> Calculating moles

Moles = mass / Molar mass

moles Fe2O3 = 177/159.69 = 1.1084 moles

moles CO = 82.8/28.01 = 2.9561 moles

⇒ CO is the limiting reactant ; Fe2O3 is the reactant in excess (there will remain 0.123 moles)

<u>Step 4</u>: Calculating moles of Fe

There will be produced 2 moles of Fe if there is consumed 3 moles of CO

If there is consumed 2.9561 moles of Fe, there is produced 1.9707 moles of Fe

<u>Step 5:</u> Calculating mass of Fe

mass = moles * Molar mass

mass of Fe = 1.9707 moles * 55.845 = 110.05 grams

The theoretical yield of solid iron is 110.05 grams

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Ulleksa [173]

<u>Answer:</u> The pH change of the buffer is 0.30

<u>Explanation:</u>

To calculate the pH of basic buffer, we use the equation given by Henderson Hasselbalch:

pOH=pK_b+\log(\frac{[\text{conjugate acid}]}{[\text{base}]})

pOH=pK_b+\log(\frac{[C_6H_5NH_3^+]}{[C_6H_5NH_2]})        .....(1)

We are given:

pK_b = negative logarithm of base dissociation constant of aniline  = 9.13

[C_6H_5NH_3^+]=0.306M

[C_6H_5NH_2]=0.418M

pOH = ?

Putting values in equation 1, we get:

pOH=9.13+\log(\frac{0.306}{0.418})\\\\pOH=8.99

To calculate pH of the solution, we use the equation:

pH+pOH=14\\pH_{initial}=14-8.99=5.01

To calculate the molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Moles hydrochloric acid solution = 0.124 mol

Volume of solution = 1 L

Putting values in above equation, we get:

\text{Molarity of HCl}=\frac{0.124}{1L}\\\\\text{Molarity of HCl}=0.124M

The chemical reaction for aniline and HCl follows the equation:

                   C_6H_5NH_2+HCl\rightarrow C_6H_5NH_3^++Cl^-

<u>Initial:</u>           0.418        0.124           0.306

<u>Final:</u>             0.294          -                0.430

Calculating the pOH by using using equation 1:

pK_b = negative logarithm of base dissociation constant of aniline  = 9.13

[C_6H_5NH_3^+]=0.430M

[C_6H_5NH_2]=0.294M

pOH = ?

Putting values in equation 1, we get:

pOH=9.13+\log(\frac{0.430}{0.294})\\\\pOH=9.29

To calculate pH of the solution, we use the equation:

pH+pOH=14\\pH_{final}=14-9.29=4.71

Calculating the pH change of the solution:

\Delta pH=pH_{initial}-pH_{final}\\\\\Delta pH=5.01-4.71=0.30

Hence, the pH change of the buffer is 0.30

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