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.
How many grams of CO will be produced?

Explanation

To solve the problem of how many grams of carbon monoxide (CO) will be produced when 12 g of carbon (C) reacts with 48 g of oxygen (O₂), we can follow these steps:

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

### Step 2: Determine the molar masses of the reactants
- Molar mass of carbon (C) = 12 g/mol
- Molar mass of oxygen (O₂) = 32 g/mol
- Molar mass of carbon monoxide (CO) = 28 g/mol (12 g/mol for C + 16 g/mol for O)

### Step 3: Calculate the number of moles of each reactant
- Moles of carbon (C):
\[ \text{Moles of C} = \frac{\text{mass of C}}{\text{molar mass of C}} = \frac{12 \text{ g}}{12 \text{ g/mol}} = 1 \text{ mol} \]

- Moles of oxygen (O₂):
\[ \text{Moles of O}_2 = \frac{\text{mass of O}_2}{\text{molar mass of O}_2} = \frac{48 \text{ g}}{32 \text{ g/mol}} = 1.5 \text{ mol} \]

### Step 4: Determine the limiting reagent
From the balanced equation, we see that:
- 2 moles of C react with 1 mole of O₂.

Thus, for 1 mole of C (which we have), we need:
\[ \text{Required moles of O}_2 = \frac{1 \text{ mol C}}{2} = 0.5 \text{ mol O}_2 \]

Since we have 1.5 moles of O₂ available, carbon (C) is the limiting reagent because it will be consumed first.

### Step 5: Calculate the amount of CO produced
According to the balanced equation:
- 2 moles of C produce 2 moles of CO.

Thus, 1 mole of C will produce 1 mole of CO. Since we have 1 mole of C:
\[ \text{Moles of CO produced} = 1 \text{ mol} \]

### Step 6: Convert moles of CO to grams
\[ \text{Mass of CO} = \text{moles of CO} \times \text{molar mass of CO} = 1 \text{ mol} \times 28 \text{ g/mol} = 28 \text{ g} \]

### Final Answer
The mass of carbon monoxide (CO) produced is **28 grams**.

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