The minimum mass of mixture of `A_(2)" and "B_(4)` required to produce at least 1 kg of each product is :
(Given At. mass of ‘A’ = 10, At mass of ‘B’ = 120)
`5A_(2)+2B_(4) rarr 2AB_(2)+4A_(2)B`

Options

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

Answer

To find the minimum mass of the mixture of \( A_2 \) and \( B_4 \) required to produce at least 1 kg of each product \( AB_2 \) and \( A_2B \), we will follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation is:
\[ 5A_2 + 2B_4 \rightarrow 2AB_2 + 4A_2B \]

### Step 2: Determine the molar masses of the reactants and products
- **Molar mass of \( A_2 \)**:
\[ \text{Molar mass of } A_2 = 2 \times \text{Atomic mass of A} = 2 \times 10 = 20 \text{ g/mol} \]

- **Molar mass of \( B_4 \)**:
\[ \text{Molar mass of } B_4 = 4 \times \text{Atomic mass of B} = 4 \times 120 = 480 \text{ g/mol} \]

- **Molar mass of \( AB_2 \)**:
\[ \text{Molar mass of } AB_2 = \text{Atomic mass of A} + 2 \times \text{Atomic mass of B} = 10 + 2 \times 120 = 250 \text{ g/mol} \]

- **Molar mass of \( A_2B \)**:
\[ \text{Molar mass of } A_2B = 2 \times \text{Atomic mass of A} + \text{Atomic mass of B} = 2 \times 10 + 120 = 140 \text{ g/mol} \]

### Step 3: Calculate the required moles of products
To produce at least 1 kg (1000 g) of each product:
- For \( AB_2 \):
\[ \text{Moles of } AB_2 = \frac{1000 \text{ g}}{250 \text{ g/mol}} = 4 \text{ moles} \]

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

### Step 4: Determine the limiting reactant
From the balanced equation:
- 2 moles of \( AB_2 \) are produced from 5 moles of \( A_2 \) and 2 moles of \( B_4 \).
- Therefore, for 4 moles of \( AB_2 \):
\[ \text{Moles of } A_2 = \frac{5}{2} \times 4 = 10 \text{ moles} \]
\[ \text{Moles of } B_4 = 2 \times 2 = 4 \text{ moles} \]

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

- **Mass of \( B_4 \)**:
\[ \text{Mass of } B_4 = 4 \text{ moles} \times 480 \text{ g/mol} = 1920 \text{ g} \]

### Step 6: Calculate the total mass of the mixture
\[ \text{Total mass} = \text{Mass of } A_2 + \text{Mass of } B_4 = 200 \text{ g} + 1920 \text{ g} = 2120 \text{ g} \]

### Final Answer
The minimum mass of the mixture of \( A_2 \) and \( B_4 \) required to produce at least 1 kg of each product is **2120 grams**.

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