Quantitative chemistry

Question 1

A student investigated the reaction of copper carbonate with dilute sulfuric acid.
Why did the balance reading decrease during the reaction?

Answer 1

A gas was lost from the flask

Question 2

A student investigated the reaction of copper carbonate with dilute sulfuric acid.
The equation for the reaction of copper carbonate is:
CuCO3 + H2SO4 –> CuSO4 + H2O + CO2
Relative formula masses: CuCO3 = 123.5, H2SO4 = 98.0, CuSO4 = 159.5
Calculate the percentage atom economy for making copper sulfate from copper carbonate

Answer 2

Atom economy = mass of wanted product
———————————
Mass of reactants
159.5 159.5
————- = ————
123.5 + 98 221.5

0.72009 x 100 = 72%

Question 3

Give one reason why it is important for the percentage atom economy of a reaction to be as high as possible

Answer 3

The higher the atom economy the less waste products are produced that need to be disposed of

Question 4

Some students were investigating the rate at which carbon dioxide gas is produced when metal carbonates react with an acid
One student reacted 1.00g of calcium carbonate with 50cm3, and excess, of dilute hydrochloric acid. The student then measured the volume of gas collected every 30 seconds
Why did the volume of gas stop changing after 210 seconds?

Answer 4

All the calcium carbonate had been used up

Question 5

Some students were investigating the rate at which carbon dioxide gas is produced when metal carbonates react with an acid
A student placed a conical flask containing 1.00g of a Group 1 carbonate (M2CO3) on a balance
He then added 50cm3, an excess, of dilute hydrochloric acid to the flask and measured the mass of carbon dioxide given off.
The equation for the reaction is:
M2CO3 + 2HCl —> 2MCl + H2O + CO2
The final mass of carbon dioxide given off was 0.32g
a) Calculate the amount, in moles, of carbon dioxide in 0.32g carbon dioxide.
Relative atomic masses: C = 12, O = 16

b) How many moles of the metal carbonate are needed to make this number of moles of carbon dioxide?

Answer 5

mass = Mr x Moles
moles = mass
——–
Mr
mass / Mr = 0.32 / 44 = 0.00727272727272…
0.00727 moles

M2CO3 : CO2
1 : 1
0.007 : 0.007
0.00727 moles