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3 years ago

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Given the balanced equation below, calculate the moles of aluminum that are needed to react completely with 28.7 moles of FeO. Y

ou must show all of your work. 2Al + 3FeO --> 3Fe +Al2O3

1 answer:

Alexxx [7]3 years ago

8 0

Answer:

43.05 moles of Al needed to react with 28.7 moles of FeO.

Explanation:

Given data:

Moles of FeO = 28.7 mol

Moles of Al needed to react with FeO = ?

Solution:

Chemical equation:

2Al + 3FeO → 3Fe + Al₂O₃

Now we will compare the moles of Al with FeO.

FeO : Al

2 : 3

28.7 : 3/2×28.7 = 43.05 mol

Thus 43.05 moles of Al needed to react with 28.7 moles of FeO.

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A 1.00 g sample of a metal X (that is known to form X ions in solution) was added to 127.9 mL of 0.5000 M sulfuric acid. After a

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<u>Answer:</u> The metal having molar mass equal to 26.95 g/mol is Aluminium

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

Molarity of NaOH solution = 0.5000 M

Volume of solution = 0.03340 L

Putting values in equation 1, we get:

$0.5000M=\frac{\text{Moles of NaOH}}{0.03340L}\\\\\text{Moles of NaOH}=(0.5000mol/L\times 0.03340L)=0.01670mol$

The chemical equation for the reaction of NaOH and sulfuric acid follows:

$2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O$

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of sulfuric acid

So, 0.01670 moles of NaOH will react with = $\frac{1}{2}\times 0.01670=0.00835mol$ of sulfuric acid

Excess moles of sulfuric acid = 0.00835 moles

Calculating the moles of sulfuric acid by using equation 1, we get:

Molarity of sulfuric acid solution = 0.5000 M

Volume of solution = 127.9 mL = 0.1279 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.5000M=\frac{\text{Moles of }H_2SO_4}{0.1279L}\\\\\text{Moles of }H_2SO_4=(0.5000mol/L\times 0.1279L)=0.06395mol$

Number of moles of sulfuric acid reacted = 0.06395 - 0.00835 = 0.0556 moles

The chemical equation for the reaction of metal (forming $M^{3+}$ ion) and sulfuric acid follows:

$2X+3H_2SO_4\rightarrow X_2(SO_4)_3+3H_2$

By Stoichiometry of the reaction:

3 moles of sulfuric acid reacts with 2 moles of metal

So, 0.0556 moles of sulfuric acid will react with = $\frac{2}{3}\times 0.0556=0.0371mol$ of metal

To calculate the molar mass of metal for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Mass of metal = 1.00 g

Moles of metal = 0.0371 moles

Putting values in above equation, we get:

$0.0371mol=\frac{1.00g}{\text{Molar mass of metal}}\\\\\text{Molar mass of metal}=\frac{1.00g}{0.0371mol}=26.95g/mol$

Hence, the metal having molar mass equal to 26.95 g/mol is Aluminium

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3 years ago

The combustion of 1.5011.501 g of fructose, C6H12O6(s)C6H12O6(s) , in a bomb calorimeter with a heat capacity of 5.205.20 kJ/°C

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Answer : The internal energy change is -2805.8 kJ/mol

Explanation :

First we have to calculate the heat gained by the calorimeter.

$q=c\times (T_{final}-T_{initial})$

where,

q = heat gained = ?

c = specific heat = $5.20kJ/^oC$

$T_{final}$ = final temperature = $27.43^oC$

$T_{initial}$ = initial temperature = $22.93^oC$

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

$q=5.20kJ/^oC\times (27.43-22.93)^oC$

$q=23.4kJ$

Now we have to calculate the enthalpy change during the reaction.

$\Delta H=-\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat gained = 23.4 kJ

n = number of moles fructose = $\frac{\text{Mass of fructose}}{\text{Molar mass of fructose}}=\frac{1.501g}{180g/mol}=0.00834mole$

$\Delta H=-\frac{23.4kJ}{0.00834mole}=-2805.8kJ/mole$

Therefore, the enthalpy change during the reaction is -2805.8 kJ/mole

Now we have to calculate the internal energy change for the combustion of 1.501 g of fructose.

Formula used :

$\Delta H=\Delta U+\Delta n_gRT$

or,

$\Delta U=\Delta H-\Delta n_gRT$

where,

$\Delta H$ = change in enthalpy = $-2805.8kJ/mol$

$\Delta U$ = change in internal energy = ?

$\Delta n_g$ = change in moles = 0 (from the reaction)

R = gas constant = 8.314 J/mol.K

T = temperature = $27.43^oC=273+27.43=300.43K$

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

$\Delta U=\Delta H-\Delta n_gRT$

$\Delta U=(-2805.8kJ/mol)-[0mol\times 8.314J/mol.K\times 300.43K$

$\Delta U=-2805.8kJ/mol-0$

$\Delta U=-2805.8kJ/mol$

Therefore, the internal energy change is -2805.8 kJ/mol

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