The volume of oxygen at STP required to burn g 4.2 of carbon completely is

Options

A. 1.12L
B. 8.96L
C. 2.24L
D. 4.48L

Explanation

To find the volume of oxygen at STP required to burn 4.2 g of carbon completely, we can follow these steps:

### Step 1: Determine the Molar Mass of Carbon
The molar mass of carbon (C) is approximately 12 g/mol.

### Step 2: Calculate the Number of Moles of Carbon
To find the number of moles of carbon in 4.2 g, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
Substituting the values:
\[ \text{Number of moles of C} = \frac{4.2 \text{ g}}{12 \text{ g/mol}} = 0.35 \text{ moles} \]

### Step 3: Write the Balanced Chemical Equation
The balanced chemical equation for the combustion of carbon is:
\[ \text{C} + \text{O}_2 \rightarrow \text{CO}_2 \]
From the equation, we see that 1 mole of carbon reacts with 1 mole of oxygen.

### Step 4: Calculate the Number of Moles of Oxygen Required
Since 0.35 moles of carbon will require an equal number of moles of oxygen:
\[ \text{Number of moles of O}_2 = 0.35 \text{ moles} \]

### Step 5: Calculate the Volume of Oxygen at STP
At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters. Therefore, the volume of 0.35 moles of oxygen is:
\[ \text{Volume of O}_2 = \text{Number of moles} \times 22.4 \text{ L/mol} \]
Substituting the values:
\[ \text{Volume of O}_2 = 0.35 \text{ moles} \times 22.4 \text{ L/mol} = 7.84 \text{ liters} \]

### Final Answer
The volume of oxygen at STP required to burn 4.2 g of carbon completely is **7.84 liters**.

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