Limestone rock contains solid calcium carbonate that can react with
acid to give products. In one experiment, 1.452 g of limestone rock is
pulverised and then treated with 25.00 mL of 1.035 M HCl solution.
After the reaction is completed, the resulting mixture is still acidic. The
excess acid then requires 15.25 mL of 0.1010 M NaOH for
neutralisation. Assuming that calcium carbonate is the only substance
in limestone rock that reacts with HCl solution:
i. Write the balanced chemical equation and net ionic equation for
the reaction that occurs between calcium carbonate and
hydrochloric acid.
[2 marks]
10
FAC 0015
ii. Calculate the number of moles of hydrochloric acid that reacts
with calcium carbonate.
[3 marks]
iii. Calculate the percent by mass of calcium carbonate in the rock.
Why did the limestone rock bring a calculator to the party? Because it wanted to do some quick calculations with acids and bases! Here are the answers to your questions:
i. The balanced chemical equation for the reaction between calcium carbonate (CaCO3) and hydrochloric acid (HCl) is:
CaCO3 + 2HCl → CaCl2 + CO2 + H2O
The net ionic equation for this reaction is:
CaCO3(s) + 2H+(aq) → Ca2+(aq) + CO2(g) + H2O(l)
ii. To calculate the number of moles of hydrochloric acid that reacts with calcium carbonate, we first need to determine the number of moles of HCl used. We can use the molarity (M) and volume (V) of the HCl solution:
Moles of HCl = Molarity × Volume
= 1.035 M × 0.02500 L
= 0.025875 moles
Since the balanced chemical equation shows a 1:1 ratio between HCl and CaCO3, the number of moles of HCl that react with CaCO3 is also 0.025875 moles.
iii. To calculate the percent by mass of calcium carbonate in the rock, we need to use the molar mass and mass of the rock sample:
Molar mass of CaCO3 = 40.08 g/mol + 12.01 g/mol + (3 × 16.00 g/mol)
= 100.09 g/mol
Mass of CaCO3 in the rock = 1.452 g - mass of impurities
Percent by mass of CaCO3 = (Mass of CaCO3 / Total mass of rock) × 100
= (mass of CaCO3 / 1.452 g) × 100
Since you didn't provide the mass of impurities, I can't calculate the exact percent by mass of calcium carbonate in the rock. But I hope these calculations help you on your chemistry journey!
i. The balanced chemical equation for the reaction between calcium carbonate (CaCO3) and hydrochloric acid (HCl) is:
CaCO3 + 2HCl -> CaCl2 + H2O + CO2
The net ionic equation for the reaction is:
CaCO3 (s) + 2H+ (aq) -> Ca2+ (aq) + H2O (l) + CO2 (g)
ii. To calculate the number of moles of hydrochloric acid that reacts with calcium carbonate, we need to use the balanced equation. From the equation, we can see that 1 mole of calcium carbonate reacts with 2 moles of hydrochloric acid.
Given:
Mass of limestone rock = 1.452 g
Volume of HCl solution = 25.00 mL
Molarity of HCl solution = 1.035 M
First, convert volume (mL) to liters (L):
25.00 mL = 0.02500 L
Now, use the formula:
moles = volume (L) x molarity
moles of HCl = 0.02500 L x 1.035 M = 0.025875 moles
Since the stoichiometric ratio is 1:2 between calcium carbonate and hydrochloric acid, the number of moles of calcium carbonate is half of the moles of hydrochloric acid.
moles of CaCO3 = 0.025875 moles / 2 = 0.0129375 moles
iii. To calculate the percent by mass of calcium carbonate in the rock, we need to compare the mass of calcium carbonate to the mass of the rock.
Given:
Mass of limestone rock = 1.452 g
moles of CaCO3 = 0.0129375 moles
The molar mass of calcium carbonate (CaCO3) is:
(40.08 g/mol) + (12.01 g/mol) + (3 x 16.00 g/mol) = 100.09 g/mol
The mass of calcium carbonate in the rock is:
mass of CaCO3 = moles of CaCO3 x molar mass of CaCO3
mass of CaCO3 = 0.0129375 moles x 100.09 g/mol = 1.294730625 g
Finally, calculate the percent by mass:
percent by mass of CaCO3 = (mass of CaCO3 / mass of limestone rock) x 100%
percent by mass of CaCO3 = (1.294730625 g / 1.452 g) x 100% = 89.15%
i. To write the balanced chemical equation for the reaction between calcium carbonate (CaCO3) and hydrochloric acid (HCl), we need to identify the products formed. It is given that the resulting mixture is still acidic after the reaction, which indicates that not all the calcium carbonate has reacted. This suggests the following equation:
CaCO3(s) + HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)
However, to write the net ionic equation, which shows only the species directly involved in the reaction, we need to represent the aqueous species as their dissociated ions:
CaCO3(s) + 2H+(aq) → Ca2+(aq) + H2O(l) + CO2(g)
ii. To calculate the number of moles of hydrochloric acid that reacts with calcium carbonate, we need to use the given information. The volume of HCl solution used is 25.00 mL, and its concentration is 1.035 M.
First, convert the volume of HCl solution used to liters:
25.00 mL = 0.02500 L
Next, use the formula:
moles of solute = concentration × volume (in liters)
moles of HCl = 1.035 M × 0.02500 L = 0.025875 moles
Since the stoichiometric ratio between HCl and CaCO3 is 1:1, the number of moles of HCl that reacts with calcium carbonate is also equal to 0.025875 moles.
iii. To calculate the percent by mass of calcium carbonate in the rock, we need to use the information about the mass of limestone rock used and the moles of calcium carbonate reacted. The mass of the limestone rock used is 1.452 g.
First, calculate the molar mass of calcium carbonate:
Molar mass of CaCO3 = (1 × atomic mass of Ca) + (1 × atomic mass of C) + (3 × atomic mass of O)
= (1 × 40.08 g/mol) + (1 × 12.01 g/mol) + (3 × 16.00 g/mol)
= 40.08 + 12.01 + 48.00
= 100.09 g/mol
Next, use the formula:
percent by mass = (moles of compound / moles of rock) × molar mass of compound × 100%
moles of rock = mass of rock / molar mass of rock
moles of rock = 1.452 g / molar mass of CaCO3
Now, substitute the values into the formula:
percent by mass of CaCO3 = (0.025875 moles / (1.452 g / 100.09 g/mol)) × 100%
percent by mass of CaCO3 = (0.025875 mol / 0.01451 mol) × 100%
Finally, calculate the percent by mass of calcium carbonate in the rock.
Limestone rock contains solid calcium carbonate that can react with
acid to give products. In one experiment, 1.452 g of limestone rock is
pulverised and then treated with 25.00 mL of 1.035 M HCl solution.
After the reaction is completed, the resulting mixture is still acidic. The
excess acid then requires 15.25 mL of 0.1010 M NaOH for
neutralisation. Assuming that calcium carbonate is the only substance
in limestone rock that reacts with HCl solution:
i. Write the balanced chemical equation and net ionic equation for
the reaction that occurs between calcium carbonate and
hydrochloric acid.
[2 marks]
10I don't understand this 10
FAC 0015 I don't understand this number either
CaCO3 + 2HCl ==> CaCl2 + CO2 + H2O
CO3^2- + 2H^+ ==> CO2 + H2O
ii. Calculate the number of moles of hydrochloric acid that reacts
with calcium carbonate.
[3 marks]
mols HCl added initially - mols HCl added in excess = mols HCl reacted with sample.
Initial mols HCl added = M HCl x L HCl = ?.
mols HCl in excess = mols NaOH x L NaOH = ?
Subtract to find mols HCl used in the reaction.
iii. Calculate the percent by mass of calcium carbonate in the rock.
Convert mols HCl used in the reaction to mols CaCO3 reacted. That is mols HCl x (1 mol CaCO3/2 mols HCl) = ?
Then grams CaCO3 = mols CaCO3 x molar mass CaCO3.
Finally, %CaCO3 = (g CaCO3/mass initial sample)*100 = ?