Mass and Volume Calculation for the Conversion of Graphite to Carbon Dioxide

To calculate the mass and volume of oxygen required to convert 2.4 kg of graphite into carbon dioxide at STP (Standard Temperature and Pressure), we need to consider the balanced chemical equation for the combustion of graphite:

C(graphite) + O2(g) -> CO2(g)

From the equation, we can see that 1 mole of graphite reacts with 1 mole of oxygen gas to produce 1 mole of carbon dioxide.

Step 1: Calculate the number of moles of graphite

Given that the mass of graphite is 2.4 kg, we need to convert it to grams before calculating the number of moles.

1 kg = 1000 g

Therefore, the mass of graphite in grams = 2.4 kg * 1000 g/kg = 2400 g

Next, we need to determine the molar mass of graphite. The molar mass of carbon (C) is 12.01 g/mol.

Molar mass of graphite = 12.01 g/mol * 1 = 12.01 g/mol

Now, we can calculate the number of moles of graphite using the formula:

Number of moles = Mass / Molar mass

Number of moles of graphite = 2400 g / 12.01 g/mol = 199.83 moles (rounded to 2 decimal places)

Step 2: Calculate the number of moles of oxygen

From the balanced chemical equation, we know that 1 mole of graphite reacts with 1 mole of oxygen gas.

Therefore, the number of moles of oxygen required is also 199.83 moles.

Step 3: Calculate the mass of oxygen

The molar mass of oxygen (O2) is 32.00 g/mol.

Mass of oxygen = Number of moles * Molar mass

Mass of oxygen = 199.83 moles * 32.00 g/mol = 6,394.56 g (rounded to 2 decimal places)

Therefore, the mass of oxygen required to convert 2.4 kg of graphite into carbon dioxide is 6,394.56 g.

Step 4: Calculate the volume of oxygen

To calculate the volume of oxygen gas at STP, we can use the ideal gas law:

PV = nRT

Where:
P = pressure (at STP, pressure = 1 atm)
V = volume
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (at STP, temperature = 273 K)

Rearranging the equation, we can solve for V:

V = nRT / P

V = 199.83 moles * 0.0821 L·atm/mol·K * 273 K / 1 atm

V = 4,924.94 L (rounded to 2 decimal places)

Therefore, the volume of oxygen required at STP to convert 2.4 kg of graphite into carbon dioxide is 4,924.94 L.