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4 years ago

Combining 0.301 mol Fe2O3

Chemistry

2 answers:

Aleksandr [31]4 years ago

4 0

The actual yield of iron (Fe) in moles is 0.207 mol.

Why?

Since we know the chemical equation, and how much Fe is produced, we can calculate the actual yield of iron in moles using the following equation:

$FeActualYield(mol)=\frac{mass(Fe)}{molarmass(Fe)}$

From the statement, we know that there are 11.6 g of Fe.

We also know the molar mass of the Fe which is:

$molarmass(Fe)=55.845\frac{g}{mol}$

Now, calculating we have:

$FeActualYield(mol)=\frac{mass(FeProduced)}{molarmass(Fe)}$

$FeActualYield(mol)=\frac{11.6g}{55.845\frac{g}{mol}}=0.207mol$

Hence, we have that the actual yield of iron (Fe) in moles is 0.207 mol.

Have a nice day!

ASHA 777 [7]4 years ago

4 0

Answer:

The actual yield of iron in moles is 34.5%

Explanation:

The balanced chemical equation is

$1Fe_2 O_3+3 C > 2 Fe + 3 CO$

The conversions are moles $Fe_2 O_3$ to moles Fe and moles Fe to mass Fe

Mole ratio of $Fe_2 O_3:Fe$ is 1 : 2

So using mole ratio we see

$0.301mol Fe_2 O_3 \times \frac {(2mol Fe)}{(1mol Fe_2 O_3 )}$

=0.601 mol Fe is produced.

By multiplying with molar mass of Fe we can convert moles Fe to mass Fe.

That is

Mass Fe = moles Fe × molar mass Fe

$=0.601mol Fe \times 55.85 g/mol$

=33.566 g Fe produced.

(Theoretical yield)

11.6 g given is the actual yield.

% yield=(actual yield )/(Theoretical yield )×100%

$= \frac {(11.6g)}{(33.6g)} \times 100$

= 34.5% is the Answer

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Explanation:

Along period:

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3 years ago

. What mass of ammonium chloride must be added to 250. mL of water to give a solution with pH

Eduardwww [97]

The mass of ammonium chloride that must be added is : ( A ) 4.7 g

Given data :

Volume of water ( V ) = 250 mL = 0.25 L

pH of solution = 4.85

Kb = 1.8 * 10⁻⁵

Kw = 10⁻¹⁴

Given that the dissolution of NH₄Cl gives NH₄⁺⁺ and Cl⁻ ions the equation is written as :

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where conc of H₃O⁺

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∴ Ka = 5.56 * 10⁻¹⁰

Next step : Determine the concentration of H₃O⁺ in the solution

pH = - log [ H₃O⁺ ] = 4.85

∴ [ H₃O⁺ ] in the solution = 1.14125 * 10⁻⁵

Next step : Determine the concentration of NH₄CI in the solution

C = [ H₃O⁺ ]² / Ka

= ( 1.14125 * 10⁻⁵ )² / 5.56 * 10⁻¹⁰

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