The minimum mass of mixture of `A_(2)` and `B_(2)` required to produce at least 1 kg of each product is : (Given At. Mass of 'A' = 10 : At. Mass of 'B' = 120 )

Options

A. 2120 gm
B. 1060 gm
C. 560 gm
D. 1660 gm

Answer

To solve the problem of finding the minimum mass of the mixture of \( A_2 \) and \( B_2 \) required to produce at least 1 kg of each product \( AB_2 \) and \( A_2B \), we can follow these steps:

### Step 1: Determine the Molar Masses
- The molar mass of \( A_2 \):
- Atomic mass of A = 10 g/mol
- Molar mass of \( A_2 = 2 \times 10 = 20 \) g/mol

- The molar mass of \( B_2 \):
- Atomic mass of B = 120 g/mol
- Molar mass of \( B_2 = 2 \times 120 = 240 \) g/mol

- The molar mass of \( AB_2 \):
- Molar mass of \( AB_2 = 10 + (2 \times 120) = 250 \) g/mol

- The molar mass of \( A_2B \):
- Molar mass of \( A_2B = (2 \times 10) + 120 = 140 \) g/mol

### Step 2: Calculate Moles Required for Each Product
- For \( AB_2 \):
- To produce 1 kg (1000 g) of \( AB_2 \):
\[ \text{Moles of } AB_2 = \frac{1000 \text{ g}}{250 \text{ g/mol}} = 4 \text{ moles} \]

- For \( A_2B \):
- To produce 1 kg (1000 g) of \( A_2B \):
\[ \text{Moles of } A_2B = \frac{1000 \text{ g}}{140 \text{ g/mol}} \approx 7.14 \text{ moles} \]

### Step 3: Determine Moles of \( A_2 \) and \( B_2 \) Required
- From the stoichiometry of the reaction:
- The reaction is \( 5 \text{ mol } A_2 + 2 \text{ mol } B_2 \rightarrow 2 \text{ mol } AB_2 + 4 \text{ mol } A_2B \).

- For 4 moles of \( AB_2 \):
- Moles of \( A_2 \) required = \( \frac{5}{2} \times 4 = 10 \text{ moles} \)
- Moles of \( B_2 \) required = \( \frac{2}{2} \times 4 = 4 \text{ moles} \)

- For 7.14 moles of \( A_2B \):
- Moles of \( A_2 \) required = \( \frac{5}{4} \times 7.14 \approx 8.925 \text{ moles} \)
- Moles of \( B_2 \) required = \( \frac{2}{4} \times 7.14 \approx 3.57 \text{ moles} \)

### Step 4: Determine the Maximum Requirement
- The maximum requirement for \( A_2 \) is 10 moles (from \( AB_2 \)).
- The maximum requirement for \( B_2 \) is 4 moles (from \( AB_2 \)).

### Step 5: Calculate the Mass of Each Component
- Mass of \( A_2 \):
\[ \text{Mass of } A_2 = 10 \text{ moles} \times 20 \text{ g/mol} = 200 \text{ g} \]

- Mass of \( B_2 \):
\[ \text{Mass of } B_2 = 4 \text{ moles} \times 240 \text{ g/mol} = 960 \text{ g} \]

### Step 6: Calculate Total Mass of the Mixture
- Total mass of the mixture:
\[ \text{Total mass} = 200 \text{ g} + 960 \text{ g} = 1160 \text{ g} \]

### Final Answer
The minimum mass of the mixture of \( A_2 \) and \( B_2 \) required to produce at least 1 kg of each product is **1160 g**.