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2 years ago

Consider the following reaction: 4Fe + 302 → 2Fe2O3. What mass of iron(III) oxide would

Chemistry

1 answer:

frutty [35]2 years ago

3 0

Answer:

Mass = 357.7 g

Explanation:

Given data:

Mass of Fe = 250 g

Mass of oxygen = 120 g

Mass of iron(III) oxide produced = ?

Solution:

Chemical equation:

4Fe + 3O₂ → 2Fe₂O₃

Number of moles of Fe:

Number of moles = mass/molar mass

Number of moles = 250 g/ 55.8 g/mol

Number of moles = 4.48 mol

Number of moles of O₂ :

Number of moles = mass/molar mass

Number of moles = 120 g/ 32 g/mol

Number of moles = 3.75 mol

Now we will compare the moles of reactants with product.

Fe : Fe₂O₃

4 : 2

4.48 : 2/4×4.48 = 2.24

O₂ : Fe₂O₃

3 : 2

3.75 : 2/3×3.75= 2.5

Less number of moles of Fe₂O₃ are produced by Fe thus it will act as limiting reactant.

Mass of Fe₂O₃:

Mass = number of moles × molar mass

Mass = 2.24 mol × 159.69 g/mol

Mass = 357.7 g

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faltersainse [42]

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2 years ago

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If you assume this reaction is driven to completion because of the large excess of one ion, what is the concentration of [Fe(SCN

viktelen [127]

Answer : The concentration of $[Fe(SCN)]^{2+}$ is, $4.32\times 10^{-4}M$

Explanation :

When we assume this reaction is driven to completion because of the large excess of one ion then we are assuming limiting reagent is $SCN^-$ and $Fe^{3+}$ is excess reagent.

First we have to calculate the moles of KSCN.

$\text{Moles of }KSCN=\text{Concentration of }KSCN\times \text{Volume of solution}$

$\text{Moles of }KSCN=0.00180M\times 0.006L=1.08\times 10^{-5}mol$

Moles of KSCN = Moles of $K^+$ = Moles of $SCN^-$ = $1.08\times 10^{-5}mol$

Now we have to calculate the concentration of $[Fe(SCN)]^{2+}$

$\text{Concentration of }[Fe(SCN)]^{2+}=\frac{\text{Moles of }[Fe(SCN)]^{2+}}{\text{Volume of solution}}$

Total volume of solution = (6.00 + 5.00 + 14.00) = 25.00 mL = 0.025 L

$\text{Concentration of }[Fe(SCN)]^{2+}=\frac{1.08\times 10^{-5}mol}{0.025L}=4.32\times 10^{-4}M$

Thus, the concentration of $[Fe(SCN)]^{2+}$ is, $4.32\times 10^{-4}M$

7 0

2 years ago

Calculate either [ H 3 O + ] or [ OH − ] for each of the solutions.

STALIN [3.7K]

Answer: Solution A : $[H_3O^+]=0.300\times 10^{-7}M$

Solution B : $[OH^-]=0.107\times 10^{-5}M$

Solution C : $[OH^-]=0.177\times 10^{-10}M$

Explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration and pOH is calculated by taking negative logarithm of hydroxide ion concentration.

$pH=-\log[H_3O^+]$

$pOH=-log[OH^-}$

$pH+pOH=14$

$[H_3O^+][OH^-]=10^{-14}$

a. Solution A: $[OH^-]=3.33\times 10^{-7}M$

$[H_3O^+]=\frac{10^{-14}}{3.33\times 10^{-7}}=0.300\times 10^{-7}M$

b. Solution B : $[H_3O^+]=9.33\times 10^{-9}M$

$[OH^-]=\frac{10^{-14}}{9.33\times 10^{-9}}=0.107\times 10^{-5}M$

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$[OH^-]=\frac{10^{-14}}{5.65\times 10^{-4}}=0.177\times 10^{-10}M$

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2 years ago

How much energy must be removed from a 94.4 g sample of benzene (molar mass= 78.11 g/mol) at 322.0 K to solidify the sample and

Kay [80]

Answer : The energy removed must be, 29.4 kJ

Explanation :

The process involved in this problem are :

$(1):C_6H_6(l)(322K)\rightarrow C_6H_6(l)(279K)\\\\(2):C_6H_6(l)(279K)\rightarrow C_6H_6(s)(279K)\\\\(3):C_6H_6(s)(279K)\rightarrow C_6H_6(s)(205K)$

The expression used will be:

$Q=[m\times c_{p,l}\times (T_{final}-T_{initial})]+[m\times \Delta H_{fusion}]+[m\times c_{p,s}\times (T_{final}-T_{initial})]$

where,

$Q$ = heat released for the reaction = ?

m = mass of benzene = 94.4 g

$c_{p,s}$ = specific heat of solid benzene = $1.51J/g^oC=1.51J/g.K$

$c_{p,l}$ = specific heat of liquid benzene = $1.73J/g^oC=1.73J/g.K$

$\Delta H_{fusion}$ = enthalpy change for fusion = $-9.8kJ/mol=-\frac{9.8\times 1000J/mol}{78g/mol}=-125.6J/g$

Now put all the given values in the above expression, we get:

$Q=[94.4g\times 1.73J/g.K\times (279-322)K]+[94.4g\times -125.6J/g]+[94.4g\times 1.51J/g.K\times (205-279)K]$

$Q=-29427.312J=-29.4kJ$

Negative sign indicates that the heat is removed from the system.

Therefore, the energy removed must be, 29.4 kJ

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2 years ago

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Basile [38]

Answer: I believe C is your best answer

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1 year ago