Preparation of cabalt Metaborate involves the following steps of reactions:
(i) `Ca_(2)B_(6)O_(11) + Na_(2)CO_(3)(aq) overset("Boiled") to CaCO_(3) "(insoluble)" + Na_(2)B_(4)O_(7) + NaBO_(2)`
(ii) `Na_(2)B_(4)O_(7) overset(Delta)to NaBO_(2) + B_(2)O_(3)`
(iii) `CoO + B_(2) O_(3) overset(Delta) to Co(BO_(2))_(2)`.
(Atomic weight: B=11, Co=59)
Mass of `Ca_(2)B_(6)O_(11)` in kg requried to produce 14.5 kg of `Co(BO_(2))_(2)`, assuming 100% yield of each reaction is

Options

A. 32.2
B. 100
C. 28.2
D. 30

Answer

To solve the problem of determining the mass of `Ca2B6O11` required to produce `14.5 kg` of `Co(BO2)2`, we will follow the steps outlined in the reactions provided.

### Step-by-Step Solution:

1. **Identify the Molar Mass of `Co(BO2)2`:**
- The molar mass of cobalt (Co) = 59 g/mol
- The molar mass of boron (B) = 11 g/mol
- The molar mass of oxygen (O) = 16 g/mol
- Molar mass of `Co(BO2)2` = 59 + 2*(11 + 16) = 59 + 2*(27) = 59 + 54 = 113 g/mol

2. **Convert the Mass of `Co(BO2)2` to Grams:**
- Given mass = 14.5 kg = 14,500 g

3. **Calculate the Number of Moles of `Co(BO2)2`:**
- Moles of `Co(BO2)2` = Given mass / Molar mass
- Moles of `Co(BO2)2` = 14,500 g / 113 g/mol ≈ 128.31 moles

4. **Determine the Moles of `Ca2B6O11` Required:**
- From the reaction steps, we see that 1 mole of `Ca2B6O11` produces 1 mole of `Co(BO2)2`.
- Therefore, moles of `Ca2B6O11` required = Moles of `Co(BO2)2` = 128.31 moles

5. **Calculate the Molar Mass of `Ca2B6O11`:**
- Molar mass of calcium (Ca) = 40 g/mol
- Molar mass of boron (B) = 11 g/mol
- Molar mass of oxygen (O) = 16 g/mol
- Molar mass of `Ca2B6O11` = 2*(40) + 6*(11) + 11*(16)
- = 80 + 66 + 176 = 322 g/mol

6. **Calculate the Mass of `Ca2B6O11` Required:**
- Mass of `Ca2B6O11` = Moles × Molar mass
- Mass of `Ca2B6O11` = 128.31 moles × 322 g/mol ≈ 41,290.42 g

7. **Convert the Mass to Kilograms:**
- Mass in kg = 41,290.42 g / 1000 = 41.29 kg

### Final Answer:
The mass of `Ca2B6O11` required to produce `14.5 kg` of `Co(BO2)2` is approximately **41.29 kg**.

---