Calcium carbonate of mass 10g was heated until there was no further change
A. Write an equation for the reaction
B. Calculate the mass of the residue
C. Calculate the volume of gas evolved at STP
D. What would be the volume of the gas measured at 15°C and 760mmHg
(C=12, O=16, Ca=40, molar mass of gas at STP is 22.4dm³)
CaCO3 ==> CaO + CO2
mols CaCO3 = g/molar mass = 10/100 = 0.1 mol
mols CaO (the residue) after heating is 0.1 mol. Convert to grams.
mols CO2 formed = 0.1 mol. You know 1 mol of any gas @ STP is 22.4 L so convert 0.1 mol to L.
Use PV = nRT to convert to other conditions.
Post your work if you get stuck.
A. Sure, here's the equation for the reaction:
CaCO3(s) -> CaO(s) + CO2(g)
B. To calculate the mass of the residue, we need to determine the mass of calcium oxide (CaO) that was formed. The molar mass of CaO is 56 g/mol, and the molar mass of CaCO3 is 100 g/mol (40 + 12 + 3 * 16).
Using the molar ratio of the reaction (1:1), we can set up a proportion to find the mass of CaO:
Mass of CaCO3 / Molar mass of CaCO3 = Mass of CaO / Molar mass of CaO
10 g / 100 g/mol = Mass of CaO / 56 g/mol
Cross-multiplying and solving for the mass of CaO:
(10 g * 56 g/mol) / 100 g/mol = 5.6 g
Therefore, the mass of the residue (CaO) is 5.6 g.
C. To calculate the volume of gas evolved at STP, we need to use the ideal gas law. The molar mass of CO2 is 44 g/mol.
First, calculate the number of moles of CO2:
Number of moles = Mass / Molar mass
Number of moles = 10 g / 44 g/mol ≈ 0.227 mol
Then, use Avogadro's law to calculate the volume at STP:
1 mol of gas occupies 22.4 dm³ at STP
Therefore, 0.227 mol of gas occupies (0.227 mol * 22.4 dm³/mol) ≈ 5.05 dm³ or 5050 cm³
So, the volume of gas evolved at STP is approximately 5.05 dm³ or 5050 cm³.
D. To calculate the volume of the gas measured at 15°C and 760 mmHg, we need to use the combined gas law. The given molar mass refers to the molar mass at STP, so we need to account for the change in conditions.
First, we need to convert the temperature to Kelvin:
Temperature in Kelvin = Temperature in Celsius + 273.15
Temperature in Kelvin = 15°C + 273.15 ≈ 288.15 K
Next, we can set up the combined gas law equation:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Where:
P1 = 760 mmHg (given)
V1 = volume of gas at STP = 5.05 dm³ = 5050 cm³ (calculated in part C)
T1 = 273.15 K (STP temperature)
P2 = pressure at 15°C and 760 mmHg
T2 = 288.15 K (15°C converted to Kelvin)
Now we can solve for V2 (volume at 15°C and 760 mmHg):
V2 = (P2 * V1 * T2) / (P1 * T1)
Substituting the values:
V2 = (760 mmHg * 5050 cm³ * 288.15 K) / (760 mmHg * 273.15 K)
Canceling similar units and simplifying:
V2 ≈ 5479 cm³
Therefore, the volume of the gas measured at 15°C and 760 mmHg would be approximately 5479 cm³.
A. The equation for the reaction between calcium carbonate (CaCO3) and heat can be written as follows:
CaCO3(s) → CaO(s) + CO2(g)
B. To calculate the mass of the residue, we need to determine the mass of the products formed. From the equation above, we see that the molar ratio between CaCO3 and CaO is 1:1. The molar mass of CaCO3 is 40 + 12 + (3 * 16) = 100 g/mol.
Therefore, if 1 mole of CaCO3 reacts to form 1 mole of CaO, then the mass of CaO formed will also be 10 g. Hence, the mass of the residue is 10 g.
C. According to the ideal gas law, at STP (Standard Temperature and Pressure), one mole of gas occupies a volume of 22.4 dm³. From the balanced equation, we see that 1 mole of CaCO3 produces 1 mole of CO2. Therefore, the volume of CO2 gas evolved at STP will also be 22.4 dm³.
D. To find the volume of the gas at 15°C and 760 mmHg, we can use the ideal gas law equation: PV = nRT.
Given:
T1 = 273 + 15 = 288 K
P1 = 760 mmHg
T2 = 273 K (STP)
P2 = 1 atm
Let's convert mmHg to atm:
P1 = 760 mmHg / 760 = 1 atm
Now we can rearrange the ideal gas law equation to find the volume at the new conditions:
V2 = (nRT2) / P2
Since we know V1 (the volume at STP) is 22.4 dm³ and n is equal to 1 mole, we can substitute these values into the equation:
V2 = (1 * 0.0821 * 288) / 1
Simplifying this, we find:
V2 = 23.63 dm³
Therefore, the volume of gas measured at 15°C and 760 mmHg is approximately 23.63 dm³.
A. The equation for the reaction of calcium carbonate (CaCO3) when heated is as follows:
CaCO3(s) → CaO(s) + CO2(g)
B. To calculate the mass of the residue, we need to determine the molar masses of CaCO3, CaO, and CO2.
Molar mass of CaCO3 = (Ca atomic mass) + (C atomic mass) + 3 * (O atomic mass)
= (40 g/mol) + (12 g/mol) + 3 * (16 g/mol)
= 40 g/mol + 12 g/mol + 48 g/mol = 100 g/mol
Since the molar mass of CaCO3 is 100 g/mol, 10 g of CaCO3 corresponds to 0.1 moles.
From the balanced chemical equation, we know that 1 mole of CaCO3 reacts to produce 1 mole of CO2. Therefore, the amount of CO2 produced is also 0.1 moles.
Molar mass of CO2 = (C atomic mass) + 2 * (O atomic mass)
= (12 g/mol) + 2 * (16 g/mol)
= 12 g/mol + 32 g/mol = 44 g/mol
Now, we can calculate the mass of CO2 produced:
Mass of CO2 = (Amount of CO2 in moles) * (Molar mass of CO2)
= 0.1 mol * 44 g/mol = 4.4 g
Therefore, the mass of the residue (CaO) is:
Mass of residue = Mass of CaCO3 - Mass of CO2 produced
= 10 g - 4.4 g = 5.6 g
C. At STP (Standard Temperature and Pressure), the volume of 1 mole of any gas is 22.4 dm³. Since 1 mole of CO2 is produced from the reaction, the volume of CO2 evolved at STP is also 22.4 dm³.
D. To calculate the volume of the gas measured at 15°C and 760 mmHg, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in this case, 760 mmHg)
V = volume (what we want to find)
n = number of moles of the gas (in this case, 0.1 moles)
R = ideal gas constant (0.0821 L•atm/mol•K)
T = temperature (in Kelvin) which is 15°C + 273.15 = 288.15 K
Rearranging the equation to solve for V, we get:
V = (nRT) / P
V = (0.1 moles * 0.0821 L•atm/mol•K * 288.15 K) / 760 mmHg
Converting mmHg to atm:
1 atm = 760 mmHg
V = (0.1 * 0.0821 * 288.15) / 1
V ≈ 2.438 L
Therefore, the volume of the gas measured at 15°C and 760 mmHg would be approximately 2.438 liters.