The reaction ` 2C + O_(2) rarr 2CO` is carried out by taking 24.0 g of carbon and 96.0 g of `O_(2)`. Find out.
How much of it is left ?

Answer

To solve the problem, we need to determine how much of the reactants (carbon and oxygen) is left after the reaction has occurred. Here’s a step-by-step breakdown of the solution:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction is:
\[ 2C + O_2 \rightarrow 2CO \]

### Step 2: Calculate the moles of the reactants
- **For Carbon (C)**:
- Given mass = 24.0 g
- Molar mass of Carbon (C) = 12.0 g/mol
- Moles of Carbon = \(\frac{\text{mass}}{\text{molar mass}} = \frac{24.0 \, \text{g}}{12.0 \, \text{g/mol}} = 2 \, \text{moles}\)

- **For Oxygen (O2)**:
- Given mass = 96.0 g
- Molar mass of Oxygen (O2) = 32.0 g/mol (16.0 g/mol for each O atom)
- Moles of Oxygen = \(\frac{96.0 \, \text{g}}{32.0 \, \text{g/mol}} = 3 \, \text{moles}\)

### Step 3: Determine the limiting reactant
From the balanced equation, we see that:
- 2 moles of Carbon react with 1 mole of Oxygen.

Now, we have:
- 2 moles of Carbon available
- 3 moles of Oxygen available

According to the stoichiometry of the reaction:
- 2 moles of Carbon would require 1 mole of Oxygen.
- Therefore, 2 moles of Carbon can react with only 1 mole of Oxygen.

### Step 4: Calculate the excess reactant
Since we have 3 moles of Oxygen but only need 1 mole to react with 2 moles of Carbon, we can find the excess:
- Moles of Oxygen left = Initial moles of Oxygen - Moles of Oxygen used
- Moles of Oxygen left = 3 moles - 1 mole = 2 moles

### Step 5: Calculate the mass of the excess Oxygen
To find the mass of the excess Oxygen:
- Mass of Oxygen left = Moles of Oxygen left × Molar mass of O2
- Mass of Oxygen left = \(2 \, \text{moles} \times 32.0 \, \text{g/mol} = 64.0 \, \text{g}\)

### Conclusion
After the reaction, 64.0 g of Oxygen (O2) is left unreacted.

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