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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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4. Why can't the subscripts be changed in a chemical equation in chemistry

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2 years ago

How many grams of aluminum oxide (Al2O3) will be formed from 10.0 grams of aluminum (Al)? 4 AL + 3 02 --> 2 AL203

d1i1m1o1n [39]

Answer:

Mass = 18.9 g

Explanation:

Given data:

Mass of Al₂O₃ formed = ?

Mass of Al = 10.0 g

Solution:

Chemical equation:

4Al + 3O₂ → 2Al₂O₃

Number of moles of Al:

Number of moles = mass/molar mass

Number of moles = 10.0 g/ 27 g/mol

Number of moles = 0.37 mol

Now we will compare the moles of Al and Al₂O₃.

Al : Al₂O₃

4 : 2

0.37 : 2/4×0.37 = 0.185 mol

Mass of Al₂O₃:

Mass = number of moles × molar mass

Mass = 0.185 mol × 101.9 g/mol

Mass = 18.9 g

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2 years ago

How do electrostatic forces affect the electrons in an atom?

svlad2 [7]

Answer:

allows the atoms of different elements to bond together to form new substances

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1 year ago

Consider the reaction below for which K = 78.2 atm-1. A(g) + B(g) ↔ C(g) Assume that 0.386 mol C(g) is placed in the cylinder re

borishaifa [10]

Answer:

1.65 L

Explanation:

The equation for the reaction is given as:

A + B ⇄ C

where;

numbers of moles = 0.386 mol C (g)

Volume = 7.29 L

Molar concentration of C = $\frac{0.386}{7.29}$

= 0.053 M

A + B ⇄ C

Initial 0 0 0.530

Change +x +x - x

Equilibrium x x (0.0530 - x)

$K = \frac{[C]}{[A][B]}$

where

K is given as ; 78.2 atm-1.

So, we have:

$78.2=\frac{[0.0530-x]}{[x][x]}$

$78.2= \frac{(0.0530-x)}{(x^2)}$

$78.2x^2= 0.0530-x$

$78.2x^2+x-0.0530=0$

Using quadratic formula;

$\frac{-b+/-\sqrt{b^2-4ac} }{2a}$

where; a = 78.2 ; b = 1 ; c= - 0.0530

= $\frac{-b+\sqrt{b^2-4ac} }{2a}$ or $\frac{-b-\sqrt{b^2-4ac} }{2a}$

= $\frac{-(1)+\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}$ or $\frac{-(1)-\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}$

= 0.0204 or -0.0332

Going by the positive value; we have:

x = 0.0204

[A] = 0.0204

[B] = 0.0204

[C] = 0.0530 - x

= 0.0530 - 0.0204

= 0.0326

Total number of moles at equilibrium = 0.0204 + 0.0204 + 0.0326

= 0.0734

Finally, we can calculate the volume of the cylinder at equilibrium using the ideal gas; PV =nRT

if we make V the subject of the formula; we have:

$V = \frac{nRT}{P}$

where;

P (pressure) = 1 atm

n (number of moles) = 0.0734 mole

R (rate constant) = 0.0821 L-atm/mol-K

T = 273.15 K (fixed constant temperature )

V (volume) = ???

$V=\frac{(0.0734*0.0821*273.15)}{(1.00)}$

V = 1.64604

V ≅ 1.65 L

3 0

2 years ago