Calculate the amount of Ni needed in the Mond's process given below
`Ni+4CO rarr Ni(CO)_(4)`
If CO used in this process is obtained through a process, in which 6 g of carbon is mixed with 4 g`CO_(2).`

Options

A. (a)14.675 g
B. (b)29 g
C. (c )58 g
D. (d)28 g

Explanation

To calculate the amount of nickel (Ni) needed in the Mond's process, we will follow these steps:

### Step 1: Determine the moles of carbon (C) and carbon dioxide (CO₂)
We are given:
- Mass of carbon (C) = 6 g
- Mass of carbon dioxide (CO₂) = 44 g

To find the number of moles, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]

The molar mass of carbon (C) is 12 g/mol and that of carbon dioxide (CO₂) is 44 g/mol.

Calculating the moles of carbon:
\[ \text{Moles of C} = \frac{6 \, \text{g}}{12 \, \text{g/mol}} = 0.5 \, \text{moles} \]

Calculating the moles of carbon dioxide:
\[ \text{Moles of CO₂} = \frac{44 \, \text{g}}{44 \, \text{g/mol}} = 1 \, \text{mole} \]

### Step 2: Determine the limiting reagent
The reaction for producing carbon monoxide (CO) from carbon and carbon dioxide is:
\[ C + CO₂ \rightarrow 2CO \]

From the stoichiometry of the reaction:
- 1 mole of C reacts with 1 mole of CO₂ to produce 2 moles of CO.

Since we have 0.5 moles of carbon and 1 mole of CO₂, carbon (C) is the limiting reagent because it will be consumed first.

### Step 3: Calculate the amount of CO produced
From the reaction:
- 1 mole of C produces 2 moles of CO.

Thus, 0.5 moles of C will produce:
\[ \text{Moles of CO} = 0.5 \, \text{moles C} \times 2 = 1 \, \text{mole of CO} \]

### Step 4: Relate the amount of CO to the amount of Ni needed
The Mond's process reaction is:
\[ Ni + 4CO \rightarrow Ni(CO)₄ \]

From the stoichiometry:
- 1 mole of Ni reacts with 4 moles of CO.

Thus, for 1 mole of CO, the amount of Ni required is:
\[ \text{Moles of Ni} = \frac{1 \, \text{mole CO}}{4} = 0.25 \, \text{moles of Ni} \]

### Step 5: Calculate the mass of Ni required
The molar mass of nickel (Ni) is approximately 58.7 g/mol. Therefore, the mass of Ni needed is:
\[ \text{Mass of Ni} = \text{moles of Ni} \times \text{molar mass of Ni} \]
\[ \text{Mass of Ni} = 0.25 \, \text{moles} \times 58.7 \, \text{g/mol} = 14.675 \, \text{g} \]

### Final Answer
The amount of nickel needed in the Mond's process is **14.675 grams**.

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