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Mass to Mass Stoichiometry Problems Answers

February 4, 2013

mass to mass stoichiometry problems answers

Introduction: Mass to mass stoichiometry problems answers

The steps involved in the ‘mass to mass conversion’ are as follows:

a)    Balance the equation for the reaction.
b)    Convert the known mass of the reactant or product to the moles of the substance.
c)    Use the balanced equation to set up the appropriate mole ratios.
d)    Use the appropriate mole ratios to calculate the number of moles of the desired reactant or product.
e)    Convert from moles back to grams if required by the problem.

Sample Mass to Mass Stoichiometry Problems Answers:

How many grams of Al can be created decomposing 9.8 g of Al2O3?

Step 1:
Balance the equation for the reaction and calculate the ratios

Al2O3 → Al + O2

2Al2O3 → 4 Al +3O2

2Al2O3 : 4Al (1:2)       

2Al2O3 : 3O2 (1:1.5)

Step 2:

Find the mass of the given

9.8 g Al2O3 are decomposed.

Step 3:

Calculate the moles of the given

1 mole of Al2O3 → 102 g of Al2O3
So, ‘x’ moles of Al
2O3 → 9.8 g of Al2O3

x = (9.8 g of Al2O3) (1 mole of Al2O3) / (102 g of Al2O3)
= 0.096 moles of Al
2O3

Step 4:

Calculate the moles using the ratios

1 mole of Al2O3 → 2 mol of Al
So, 0.096 moles of Al
2O3 → ‘y’ mol of Al

y = (0.096 moles of Al2O3) (2 mol of Al)/ (1 mole of Al2O3)
= 0.19 mol of Al

Step 5:

Calculate the mass using the new moles

1 mol of Al → 27 g of Al
So, 0.19 mol of Al → ‘z’ g of Al

z = (0.19 mol of Al) (27 g of Al) / (1 mol of Al)
= 5.1 g of Al

Final answer: 5.1 g of Al

More Mass to Mass Stoichiometry Problems Answers: Problem 1

Ca3 (PO4)2 + 6 SiO2 + 10 C → 6 CaSiO3 + 10 CO + P4a)    Determine the mass of SiO2 consumed if 1.8 g of carbon is consumed. b)    Determine the mass of P4 produced if 1.8 g of carbon is consumed. 


c)    Determine the mass of carbon consumed if 12.4 g of P4 is produced.

Solution:

We need to balance this equation:

Ca3 (PO4)2 + 6 SiO2 + 10 C → 6 CaSiO3 + 10 CO + P4

Now, 1.8 g of carbon is consumed in a) and b).

We know that 1 mol of C 12 g of C
So, ‘x’ moles of C
1.8 g of C

x = (1.8 g of C) (1 mol of C) / (12 g of C)
= 0.15 moles of C

a) Now the mole ratio of SiO2 to C is 6:10 = 3:5

3 moles of SiO2 5 moles of C
‘y’ moles of SiO
2 0.15 moles of C

y = (0.15 moles of C) (3 moles of SiO2) / (5 moles of C)
= 0.09 moles of SiO
2

Molar mass of SiO2 is 60.08 g/mol.
1 mol of SiO
2 60.08 g of SiO2
0.09 moles of SiO
2 ‘z’ g of SiO2

z = (0.09 moles of SiO2) (60.08 g of SiO2) / (1 mol of SiO2)
= 5.41 g of SiO
2
So, the answer to a) is 5.41 g of SiO2

b) Now the mole ratio of P4 to C is 1:10

1mole of P4 10 moles of C
‘p’ moles of P
4 0.15 moles of C

p = (0.15 moles of C) (1 mole of P4) / (10 moles of C)
= 0.015 moles of P
4

Molar mass of P4 is 123.895 g/mol.
1 mol of P
4 123.895 g of P4
0.015 moles of P
4 ‘q’ g of P4

z = (0.015 moles of P4) (123.895g of P4) / (1 mol of P4)
= 1.86 g of P
4
So, the answer to b) is 1.86 g of P4

c) 12.4 g of P4

1 mole of P4 123.895 g of P4
‘a’ moles of P
4 12.4 g of P4

a = (12.4 g of P4) (1 mole of P4) / (123.895 g of P4)

a = 0.1 moles of P4

The mole ratio of P4 to C is 1:10

1mole of P4 10 moles of C
0.1 moles of P
4 ‘b’ moles of C

b = (0.1 moles of P4) (10 moles of C)/ (1 mole of P4)
= 1 mole of C
1 mole of C
24 g of C

So, the answer to c) is 12g of C

Mass to Mass Stoichiometry Problems Answers: Problem 2

Solid lithium hydroxide is used in space vehicles to remove exhaled carbon dioxide from the living environment by forming solid lithium carbonate and liquid water. What mass of gaseous carbon dioxide can be absorbed by 1.00 kg of lithium hydroxide?

Solution:

From the description, the equation can be figured out:
LiOH (s) +CO2 (g) → Li2CO3 (s) + H2O (l)

After balancing,

2 LiOH (s) +CO2 (g) → Li2CO3 (s) + H2O (l)

Molar mass of LiOH is 23.95 g/mol

Now, 1 mole → 23.95 g

So, ‘x’ moles → 1kg = 1000 g

x = (1000 g) (1 mol) / (23.95 g)
= 41.8 mol of LiOH

Now, from the equation, 2 moles of LiOH yields 1 mol of CO2. So, the mole ratio of LiOH to CO2 is 2:1.

2 mol of LiOH → 1 mol of CO2
So, 41.8 mol of LiOH → ‘y’ mol of CO
2

y = (41.8 mol of LiOH) (1 mol of CO2) / (2 mol of LiOH)
= 20.9 mol of CO
2

Molar mass of CO2 is 44 g/mol
1mol of CO
2 → 44 g of CO2
So, 20.9 mol of CO
2 → ‘z’ g of CO2

z = (20.9 mol of CO2) (44 g of CO2) / (1mol of CO2)
= 919.6
= 9.19 x 10 ^2 g of CO
2
≈9.2 x 10^2 g of CO
2
= 920 g of CO
2

Thus, 920 g of CO2 is our final answer.

Mass to Mass Stoichiometry Problems Answers: Problem 3

Over the years, the thermite reaction has been used for welding railroad rails, in incendiary bombs, and to ignite solid-fuel rocket motors. The reaction is

Fe2O3 (s) + 2 Al (s) → 2 Fe (l) + Al2O3 (s)

a)    What is the mass of iron (III) oxide that must be used to produce 15 g of iron?
b)    What is the mass of aluminum that must be used to produce 15 g of iron?
c)    What is the maximum mass of aluminum oxide that could be produced?

Solution:

The reaction is already balanced.
Fe2O3 (s) + 2 Al (s) → 2 Fe (l) + Al2O3 (s)
15 g of iron is produced.

The molar mass of iron is 55.847 g/mol

That implies, 1mole → 55.847 g
‘x’ moles → 15 g

x = (15 g) (1 mol) / (55.847 g)
= 0.27 mol

a)
Now, the mole ratio of iron (III) oxide to iron is 1:2.

1 mol Fe2O3 → 2 mol of Fe
‘y’ mol of Fe
2O3 → 0.27 mol of Fe

y = (0.27 mol of Fe) (1 mol Fe2O3) / (2 mol of Fe)
= 0.135 mol of Fe
2O3

The molar mass of Fe2O3 is 159.692 g/ mol

1 mol of Fe2O3 → 159.692 g of Fe2O3
So, 0.135 mol of Fe
2O3 → ‘z’ of Fe2O3

z = (0.135 mol of Fe2O3) (159.692 g of Fe2O3) / (1 mol of Fe2O3)
= 21.558 g of Fe
2O3
= 21.56 g of Fe
2O3

Therefore, the mass of the iron (III) oxide that must be used to produce 15 g of Fe is 21.56 g of Fe2O3.

b) Now, the mole ratio of aluminum to iron is 2:2 = 1:1.

1 mole of Al → 1 mole of Fe
So, ‘p’ mol of Al → 0.27 mol of Fe

p = (0.27 mol of Fe) (1 mole of Al) / (1 mole of Fe)
= 0.27 mol of Al

Molar mass of Al is 26.92 g/mol

1 mol of Al → 26.92 g of Al

0.27 mol of Al → ‘q’ g of Al

q = (0.27 mol of Al) (26.92 g of Al) / (1 mol of Al)
= 7.27 g of Al

Therefore the mass of Al required to produce 15 g of Fe is 7.27 g.


c)
The mole ratio of Al2O3 to Fe is 1:2.

1 mole of Al2O3 → 2 moles of Fe
‘a’ moles of Al
2O3 → 0.27 moles of Fe

a = (0.27 mole of Fe) (1 mole of Al2O3) / (2 moles of Fe)

= 0.135 moles of Al2O3

1 mole of Al2O3 → 101.96 g of Al2O3
0.135 moles of Al
2O → ‘b’ g of Al2O3

b = (0.135 moles of Al2O3) (101.96 g of Al2O3) / (1 mole of Al2O3)
= 13.76 g of Al
2O3

So, the final answer to the c) option is 13.76 g of Al2O3.

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