# 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 Al _{2}O_{3}?**

Step 1:

Balance the equation for the reaction and calculate the ratios

**Al _{2}O_{3} → Al + O_{2}**

2Al_{2}O_{3} → 4 Al +3O_{2}

**2Al**_{2}**O**_{3}** : 4Al (1:2) **

**2Al**_{2}**O**_{3}** : 3O**_{2}** (1:1.5)**

Step 2:

Find the mass of the given

**9.8 g Al**_{2}**O**_{3}** are decomposed.**

Step 3:

Calculate the moles of the given

**1 mole of Al**_{2}**O**_{3}** → 102 g of Al**_{2}**O**_{3}

So, ‘x’ moles of Al_{2}**O**_{3}** → 9.8 g of Al**_{2}**O**_{3}

x = (9.8 g of Al_{2}**O**_{3}**) (1 mole of Al**_{2}**O**_{3}**) / (102 g of Al**_{2}**O**_{3}**)
= 0.096 moles of Al**

_{2}

**O**

_{3}

Step 4:

Calculate the moles using the ratios

**1 mole of Al**_{2}**O**_{3}** → 2 mol of Al
So, 0.096 moles of Al**

_{2}

**O**

_{3}

**→ ‘y’ mol of Al**

y = (0.096 moles of Al_{2}**O**_{3}**) (2 mol of Al)/ (1 mole of Al**_{2}**O**_{3}**)
= 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**

**Ca _{3} (PO_{4})_{2} + 6 SiO_{2} + 10 C → 6 CaSiO_{3} + 10 CO + P_{4}**

**a) Determine the mass of SiO**

_{2}consumed if 1.8 g of carbon is consumed.

**b) Determine the mass of P**_{4}produced if 1.8 g of carbon is consumed.

c) Determine the mass of carbon consumed if 12.4 g of P_{4} is produced.

Solution:

We need to balance this equation:

**Ca _{3} (PO_{4})_{2} + 6 SiO_{2} + 10 C → 6 CaSiO_{3} + 10 CO + P_{4}**

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 **SiO**_{2}** to C is 6:10 = 3:5**

**3 moles of SiO**_{2}** ****→**** 5 moles of C
‘y’ moles of SiO**

_{2}

**→**

**0.15 moles of C**

y = (0.15 moles of C) (3 moles of SiO_{2}**) / (5 moles of C)
= 0.09 moles of SiO**

_{2}

Molar mass of SiO_{2}** is 60.08 g/mol.
1 mol of SiO**

_{2 }

**→**

**60.08 g of SiO**

_{2}

0.09 moles of SiO

0.09 moles of SiO

_{2}

**→**

**‘z’ g of SiO**

_{2}

**z = (0.09 moles of SiO**_{2}**) (60.08 g of SiO**_{2}**) / (1 mol of SiO**_{2}**)
= 5.41 g of SiO**

_{2}

**So, the answer to a) is 5.41 g of SiO**

_{2}**b)** Now the mole ratio of **P**_{4}** to C is 1:10 **

**1mole of P**_{4}** →**** 10 moles of C
‘p’ moles of P**

_{4}

**→**

**0.15 moles of C**

p = (0.15 moles of C) (1 mole of P_{4}**) / (10 moles of C)
= 0.015 moles of P**

_{4}

Molar mass of P_{4}** is 123.895 g/mol.
1 mol of P**

_{4}

**→**

**123.895 g of P**

_{4}

0.015 moles of P

0.015 moles of P

_{4}

**→**

**‘q’ g of P**

_{4}

z = (0.015 moles of P_{4}**) (123.895g of P**_{4}**) / (1 mol of P**_{4}**)
= 1.86 g of P**

_{4}

**So, the answer to b) is 1.86 g of P4**

**c) ****12.4 g of P**_{4}

**1 mole of P**_{4}**→**** 123.895 g of P**_{4}

‘a’ moles of P_{4}** ****→**** 12.4 g of P**_{4}

a = (12.4 g of P_{4}**) (1 mole of P**_{4}**) / (123.895 g of P**_{4}**)**

a = 0.1 moles of P_{4}

The mole ratio of P_{4 }**to C is 1:10**

1mole of P_{4 }**→**** 10 moles of C
0.1 moles of P**

_{4}

**→**

**‘b’ moles of C**

b = (0.1 moles of P_{4}**) (10 moles of C)/ (1 mole of P**_{4}**)
= 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)}** +CO**_{2 (g)}** → Li _{2}CO**

_{3 (s)}

**+ H**

_{2}

**O**

_{(l)}

After balancing,

2 LiOH _{(s)}** +CO**_{2 (g)}** → Li _{2}CO**

_{3 (s)}

**+ H**

_{2}

**O**

_{(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 CO**_{2}**. So, the mole ratio of LiOH to CO**_{2}** is 2:1.**

2 mol of LiOH → 1 mol of CO_{2}

So, 41.8 mol of LiOH → ‘y’ mol of CO_{2}

y = (41.8 mol of LiOH) (1 mol of CO_{2}**) / (2 mol of LiOH)
= 20.9 mol of CO**

_{2}

Molar mass of CO_{2}** is 44 g/mol
1mol of CO**

_{2}

**→ 44 g of CO**

_{2}

So, 20.9 mol of CO

So, 20.9 mol of CO

_{2}

**→ ‘z’ g of CO**

_{2}

z = (20.9 mol of CO_{2}**) (44 g of CO**_{2}**) / (1mol of CO**_{2}**)
= 919.6
= 9.19 x 10 ^2 g of CO**

_{2}

≈9.2 x 10^2 g of CO

≈9.2 x 10^2 g of CO

_{2}

= 920 g of CO

= 920 g of CO

_{2}

Thus, 920 g of CO

Thus, 920 g of CO

_{2}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

**Fe _{2}O_{3 (s)} + 2 Al _{(s)} → 2 Fe _{(l)} + Al_{2}O_{3} _{(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.

**Fe**_{2}**O**_{3 (s)}** + 2 Al **_{(s)}** → 2 Fe **_{(l)}** + Al**_{2}**O**_{3 (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 m

**ole ratio of iron (III) oxide to iron is 1:2**.

**1 mol Fe**_{2}**O**_{3}** → 2 mol of Fe
‘y’ mol of Fe**

_{2}

**O**

_{3}

**→ 0.27 mol of Fe**

y = (0.27 mol of Fe) (1 mol Fe_{2}**O**_{3}**) / (2 mol of Fe)
= 0.135 mol of Fe**

_{2}

**O**

_{3}

The molar mass of Fe_{2}**O**_{3}** is 159.692 g/ mol**

1 mol of Fe_{2}**O**_{3}** → 159.692 g of Fe**_{2}**O**_{3}

So, 0.135 mol of Fe_{2}**O**_{3}** → ‘z’ of Fe**_{2}**O**_{3}

z = (0.135 mol of Fe_{2}**O**_{3}**) (159.692 g of Fe**_{2}**O**_{3}**) / (1 mol of Fe**_{2}**O**_{3}**)
= 21.558 g of Fe**

_{2}

**O**

_{3}

= 21.56 g of Fe

= 21.56 g of Fe

_{2}

**O**

_{3}

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

_{2}O_{3}.**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 Al**

_{2}

**O**

_{3}

**to Fe is 1:2.**

**1 mole of Al**_{2}**O**_{3}** → 2 moles of Fe
‘a’ moles of Al**

_{2}

**O**

_{3}

**→ 0.27 moles of Fe**

a = (0.27 mole of Fe) (1 mole of Al_{2}**O**_{3}**) / (2 moles of Fe)**

= 0.135 moles of Al_{2}**O**_{3}

1 mole of Al_{2}**O**_{3}** → 101.96 g of Al**_{2}**O**_{3}

0.135 moles of Al_{2}**O**_{3 }** → ‘b’ g of Al**_{2}**O**_{3}

b = (0.135 moles of Al_{2}**O**_{3}**) (101.96 g of Al**_{2}**O**_{3}**) / (1 mole of Al**_{2}**O**_{3}**)
= 13.76 g of Al**

_{2}

**O**

_{3}

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

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

_{2}O_{3.}