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 many grams of the other reactant should be taken so that nothing is left at the end of the reaction ?

Answer

To solve the problem, we need to determine how many grams of carbon should be taken so that all the oxygen reacts completely without any leftover reactants.

### Step-by-Step Solution:

1. **Write the Balanced Equation**:
The balanced chemical reaction is:
\[ 2C + O_2 \rightarrow 2CO \]
This indicates that 2 moles of carbon (C) react with 1 mole of oxygen (O₂) to produce 2 moles of carbon monoxide (CO).

2. **Calculate Moles of Reactants**:
- Given mass of carbon (C) = 24.0 g
- Molar mass of carbon = 12 g/mol
\[ \text{Moles of } C = \frac{24.0 \text{ g}}{12 \text{ g/mol}} = 2 \text{ moles} \]

- Given mass of oxygen (O₂) = 96.0 g
- Molar mass of oxygen = 32 g/mol
\[ \text{Moles of } O_2 = \frac{96.0 \text{ g}}{32 \text{ g/mol}} = 3 \text{ moles} \]

3. **Determine the Limiting Reagent**:
According to the balanced equation, 2 moles of carbon react with 1 mole of oxygen.
- For 3 moles of O₂, the required moles of carbon can be calculated as follows:
\[ \text{Required moles of } C = 2 \text{ moles of } C \times 3 \text{ moles of } O_2 = 6 \text{ moles of } C \]
Since we only have 2 moles of carbon, carbon (C) is the limiting reagent.

4. **Calculate the Required Mass of Carbon**:
To completely react with 3 moles of O₂, we need 6 moles of carbon. Now, we find the mass of 6 moles of carbon:
\[ \text{Mass of 6 moles of } C = 6 \text{ moles} \times 12 \text{ g/mol} = 72 \text{ g} \]

5. **Determine Additional Carbon Needed**:
We initially have 24 g of carbon. To find out how much more carbon is needed:
\[ \text{Additional carbon needed} = 72 \text{ g} - 24 \text{ g} = 48 \text{ g} \]

### Final Answer:
To ensure that nothing is left at the end of the reaction, **48 grams of carbon should be added**.

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