Calculate the mass of CaCO3 that would produce 40 mL at STP.

STP means P= 1 atm and T= 0 degrees C.

How do I find the mass?

To find the mass of CaCO3 that would produce 40 mL at STP, you can follow these steps:

1. Convert the given volume from mL to liters:
40 mL = 40/1000 = 0.04 L

2. Use the ideal gas law equation PV = nRT to calculate the number of moles (n) of CO2 gas produced. At STP conditions (P = 1 atm and T = 0 degrees C), the values are:
P = 1 atm
V = 0.04 L (from step 1)
R = 0.0821 L·atm/(mol·K) (ideal gas constant)
T = 273.15 K (0 degrees C + 273.15)

Rearrange the equation to solve for n:
n = PV / RT

Substitute the given values:
n = (1 atm) * (0.04 L) / (0.0821 L·atm/(mol·K) * 273.15 K)

3. Use the balanced chemical equation for the decomposition of CaCO3 to determine the stoichiometry between CaCO3 and CO2.
CaCO3(s) → CaO(s) + CO2(g)

The coefficient for CaCO3 is 1, meaning that 1 mole of CaCO3 produces 1 mole of CO2 gas.

4. Multiply the moles of CO2 gas by the molar mass of CaCO3 to find the mass:
Molar mass of CaCO3 = mass of 1 mole of Ca + mass of 1 mole of C + mass of 3 moles of O

Mass of Ca = 40.08 g/mol
Mass of C = 12.01 g/mol
Mass of O = 16.00 g/mol

Molar mass of CaCO3 = (40.08 g/mol) + (12.01 g/mol) + (3 * 16.00 g/mol)

Multiply the moles of CO2 gas by the molar mass of CaCO3:
Mass = n * Molar mass of CaCO3

Substitute the calculated value for n from step 2 and the molar mass of CaCO3:
Mass = (n) * [(40.08 g/mol) + (12.01 g/mol) + (3 * 16.00 g/mol)]

By following these steps, you can calculate the mass of CaCO3 that would produce 40 mL of CO2 gas at STP.

To find the mass of CaCO3 that would produce 40 mL at STP, you need to follow these steps:

1. Convert the given volume (40 mL) to liters. Since 1 L is equal to 1000 mL, divide 40 mL by 1000 to get the volume in liters.

40 mL ÷ 1000 = 0.04 L

2. Use the ideal gas law equation, PV = nRT, to calculate the number of moles (n) of CO2 gas produced. At STP, the pressure (P) is 1 atm, the temperature (T) is 0°C (273.15 K), and the volume (V) is 0.04 L. The gas constant (R) is 0.0821 L·atm/mol·K.

PV = nRT

1 atm (0.04 L) = n (0.0821 L·atm/mol·K) (273.15 K)

Solve for n (moles of CO2):

n = (1 atm) (0.04 L) / (0.0821 L·atm/mol·K) (273.15 K)

3. Convert the moles of CO2 to moles of CaCO3. The balanced chemical equation for the reaction of CaCO3 to produce CO2 gas is:

CaCO3 → CaO + CO2

The stoichiometry of the reaction is 1 mole of CaCO3 produces 1 mole of CO2.

Therefore, the number of moles of CaCO3 is the same as the number of moles of CO2:

moles of CaCO3 = moles of CO2

4. Convert moles of CaCO3 to grams. The molar mass of CaCO3 is the sum of the atomic masses of one calcium atom (Ca), one carbon atom (C), and three oxygen atoms (O).

The atomic masses (rounded to two decimal places) are:
Ca: 40.08 g/mol
C: 12.01 g/mol
O: 16.00 g/mol

Molar mass of CaCO3 = (40.08 g/mol) + (12.01 g/mol) + (3 * 16.00 g/mol)

Calculate the mass of CaCO3 using the moles of CaCO3 calculated in step 3:

mass of CaCO3 = moles of CaCO3 × molar mass of CaCO3

By following these steps, you can find the mass of CaCO3 that would produce 40 mL at STP.

Use PV = nRT.

You know P, V, R, and T. Don't forget to change T to Kelvin (273 + C = K). Use V in liters. Solve for n. Then n = # mols = grams/molar mass.