Carbon burns in oxygen as shown by the chemical equation: `2C(s) + O_(2)(g) to 2CO(g)`.
A reaction is carried out starting with 12 g carbon and 48 g oxygen.
What mass of the limiting reactant should be taken so that the end product is only CO?

Explanation

To solve the problem, we need to determine how much carbon (the limiting reactant) is required to completely react with the given amount of oxygen (48 g) to produce carbon monoxide (CO). We will follow these steps:

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

### Step 2: Calculate the number of moles of carbon and oxygen
- **For Carbon (C)**:
- Given mass = 12 g
- Molar mass of carbon = 12 g/mol
\[ \text{Number of moles of carbon} = \frac{\text{mass}}{\text{molar mass}} = \frac{12 \text{ g}}{12 \text{ g/mol}} = 1 \text{ mole} \]

- **For Oxygen (O2)**:
- Given mass = 48 g
- Molar mass of O2 = 32 g/mol
\[ \text{Number of moles of oxygen} = \frac{48 \text{ g}}{32 \text{ g/mol}} = 1.5 \text{ moles} \]

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

In our case:
- We have 1 mole of carbon and 1.5 moles of oxygen.
- To completely react with 1 mole of oxygen, we would need 2 moles of carbon. Since we only have 1 mole of carbon, carbon is the limiting reactant.

### Step 4: Calculate the mass of carbon needed to react with 48 g of oxygen
To find out how much carbon is required to react with 48 g of oxygen, we can set up a proportion based on the stoichiometry of the reaction:
- From the balanced equation:
- 1 mole of O2 (32 g) reacts with 2 moles of C (24 g).

Let \( x \) be the mass of carbon that will react with 48 g of oxygen:
\[ \frac{x \text{ g}}{24 \text{ g}} = \frac{48 \text{ g}}{32 \text{ g}} \]

### Step 5: Solve for \( x \)
Cross-multiplying gives us:
\[ 32x = 48 \times 24 \]
Calculating the right side:
\[ 32x = 1152 \]
Now, divide both sides by 32:
\[ x = \frac{1152}{32} = 36 \text{ g} \]

### Conclusion
The mass of the limiting reactant (carbon) that should be taken to ensure that the end product is only carbon monoxide (CO) is **36 g**.

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