lab 217074 1280 Stoichiometry Questions and Answers(Page 1)

Stoichiometry Questions and Answers(Page 1)

Stoichiometry Questions and Answers (Page 1)


Use the values of Ar which follow to answer the questions
below.
[H = 1; C = 12; N = 14; O = 16; Ne = 20; Na = 23;
Mg = 24; S = 32; K = 39; Fe = 56; Cu = 63.5; Zn = 65.
]

One mole of any gas at rtp occupies 28 dm3.



Question 1

Calculate the number of moles in:
a. 2 g of neon atoms
b. 4 g of magnesium atoms
c. 24g of carbon atoms.

Solution 1

a. Number of moles = mass ÷ molar mass
= 2/20
= 0.1 moles

b. Number of moles = mass ÷ molar mass
= 4/24
= 0.167 moles

c. Number of moles = mass ÷ molar mass
= 24/12
= 2 moles




Question 2

Calculate the mass of:
a. 0.1 mole of oxygen molecules
b. 5 moles of sulfur atoms
c. 0.25 mole of sodium atoms

Solution 2

a. each oxygen molecule contains 2 atoms of oxygen. So since the Mr of an oxygen atom is 16, the molar mass of oxygen molecules is 2 × 16 = 32g
Mass = Number of moles × molar mass
= 0.1 × 32
= 3.2 g

b. Mass = Number of moles × molar mass
= 5 × 32
= 160g

c. Mass = Number of moles × molar mass
= 0.25 × 23
= 5.75g



Question 3

Use the following values of Ar to answer the questions below. [H = 1; C = 12; O = 16; Ca = 40.] a. Determine the empirical formula of an oxide of calcium formed when 0.4g of calcium reacts with 0.16g of oxygen.
b. Determine the empirical formula of an organic hydrocarbon compound which contains 80% by mass of carbon and 20% by mass of hydrogen. If the Mr of the compound is 30, what is its molecular formula?

Solution 3

a.

Calcium Oxygen
Mass 0.4 0.16
Molar mass 40 16
Number of moles 0.4/40 = 0.01 0.16/16 = 0.01
Molar ratio 1 1

So the empirical formula is CaO.

b.

Carbon Hydrogen
Mass 80 20
Molar mass 12 1
Number of moles 80/12 = 6.67 20/1 = 20
Molar ratio 6.67/6.67 = 1 20/6.67 = 3

So the empirical formula is CH3

To find molecular formula we multiply empirical formula by number of formula units.

Formula units = Molecular mass ÷ Empirical mass
= 30 ÷ (12 + 3)
= 30 ÷ 15
= 2

Molecular mass = 2(CH3)
= C2H6




Question 4

Use the data in the table below to answer the questions which follow.

Element Ar
H 1
C 12
N 14
O 16
Na 23
Mg 24
Si 28
S 32
Cl 35.5
Fe 56

Calculate the mass of:
a. 1 mole of:
(i) chlorine molecules
(ii) iron(iii) oxide.
b. 0.5 mole of magnesium nitrate

Solution 4

a.(i) 35.5g
(ii) iron(iii) oxide is Fe2O3
mass of 1 mole = (56 × 2 + 16 × 3) = 160g

b.(i) molar mass of Mg(NO3)2
= 24 + 14 × 2 + 16 × 6
= 148g
therefore mass of 0.5 moles = 0.5 × 148
= 74g



Question 5

NB the molar gas molar varies according to local conditions, as such, for cambridge exams its 24 dm3 and for zimsec exams its 28 dm3.
Calculate the volume occupied, at rtp, by the following gases.
a. 12.5 moles of sulfur dioxide gas.
b. 0.15 mole of nitrogen gas.

Solution 5

a. Volume = number of moles × molar gas volume
= 12.5 × 28 dm3
= 350 dm3

b. Volume = number of moles × molar gas volume
= 0.15 × 28 dm3
= 4.2 dm3




Question 6

Calculate the number of moles of gas present in the following:
a. 36cm3 of sulfur dioxide
b. 144 dm3 of hydrogen sulfide.

Solution 6

a. since the volume is given in cm3 we can as well first convert the molar gas volume to cm3. 28 dm3 = 28 000 cm3.
Number of moles = volume ÷ molar volume
= 36/28 000
= 0.0013 moles

b. Number of moles = volume ÷ molar volume
= 144/28
= 5.14 moles



Question 7

Use the following experimental information to determine the empirical formula of an oxide of silicon.

Mass of crucible = 18.20 g
Mass of crucible + silicon = 18.48 g
Mass of crucible + oxide of silicon = 18.80 g

Solution 7

To find the empirical formula of a compound we need the masses of the elements that make up the compound.

Mass of silicon = 18.48 – 18.20 = 0.28g
Mass of oxygen = 18.80 – 18.48 = 0.32g

Silicon Oxygen
Mass 0.28 0.32
Molar mass 28 32
Number of moles 0.28/28 = 0.01 0.32/16 = 0.02
Molar ratio 0.01/0.01 = 1 0.02/0.01 = 2

Empirical formula = SiO2



Similar Posts

4 Comments

  1. Thank you for the sensible critique. Me & my neighbor were just preparing to do some research on this. We got a grab a book from our area library but I think I learned more clear from this post. I am very glad to see such magnificent info being shared freely out there.

  2. Just wish to say this blog is very good. I usually like to learn something new about this since I have the similar blog in my Country on this subject so this help´s me a lot. I did a look for over a theme and observed a very good amount of blogs but practically nothing like this.Thanks for sharing so significantly within your blog.

  3. Many thanks pertaining to spreading the following superb written content on your website. I noticed it on the internet. I will check back again when you publish additional aricles.

  4. Pretty good article. I just stumbled upon your blog and wanted to say that I have really enjoyed reading your blog posts. Any way Ill be coming back and I hope you post again soon.

Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.