Calculation of Volume of CO2 Obtained from Heating MgCO3

To calculate the volume of CO2 obtained from heating 16.8g MgCO3 with a percent yield of 75%, we need to follow the steps below:

Step 1: Write the Balanced Chemical Equation

MgCO3(s) → MgO(s) + CO2(g)

Step 2: Calculate the Number of Moles of MgCO3

Molar mass of MgCO3 = 24.31 + 12.01 + (3 x 16.00) = 84.31 g/mol

Number of moles of MgCO3 = Mass ÷ Molar mass
= 16.8 g ÷ 84.31 g/mol
= 0.1992 mol

Step 3: Calculate the Theoretical Yield of CO2

From the balanced chemical equation, we can see that one mole of MgCO3 produces one mole of CO2. Therefore, the theoretical yield of CO2 is also 0.1992 mol.

Step 4: Calculate the Actual Yield of CO2

Actual yield = Percent yield x Theoretical yield
= 75% x 0.1992 mol
= 0.1494 mol

Step 5: Calculate the Volume of CO2

We can use the ideal gas law to calculate the volume of CO2.

PV = nRT

Where:
P = Pressure (atm)
V = Volume (L)
n = Number of moles (mol)
R = Gas constant (0.0821 L atm/mol K)
T = Temperature (K)

Assuming standard temperature and pressure (STP) of 0°C (273 K) and 1 atm, we can calculate the volume of CO2 as follows:

V = nRT/P
= 0.1494 mol x 0.0821 L atm/mol K x 273 K / 1 atm
= 3.52 L

Therefore, the volume of CO2 obtained from heating 16.8g MgCO3 with a percent yield of 75% is 3.52 L.