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 moles of CO will be produced?

Explanation

To determine how many moles of CO will be produced from the reaction of 12 g of carbon and 48 g of oxygen, we can follow these steps:

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

### Step 2: Calculate the molar masses
- 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

### Step 3: Convert grams of reactants to moles
- 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):
\[ \text{Moles of O2} = \frac{\text{mass of O2}}{\text{molar mass of O2}} = \frac{48 \, \text{g}}{32 \, \text{g/mol}} = 1.5 \, \text{mol} \]

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

Since we have 1 mole of C, we would need:
\[ \text{Required moles of O2} = \frac{1 \, \text{mol C}}{2} = 0.5 \, \text{mol O2} \]

Since we have 1.5 moles of O2 available, carbon (C) is the limiting reagent.

### Step 5: Calculate the moles of CO produced
From the balanced equation, 2 moles of C produce 2 moles of CO. Therefore, 1 mole of C will produce:
\[ \text{Moles of CO} = 1 \, \text{mol C} \rightarrow 1 \, \text{mol CO} \]

### Conclusion
Thus, the number of moles of CO produced from the reaction is:
\[ \text{Moles of CO} = 1 \, \text{mol} \]

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