`BCl_(3)+H_(2)+Coverset(1700-1800^(@)C)to`Product(s)
The product(s) of above reaction is/are

Options

A. `C Cl_(4)+B_(2)H_(6)`
B. `B_(4)C+HCl`
C. `B+Cl_(2)+CH_(4)`
D. `B_(2)H_(6)+CHCl_(3)`

Explanation

To solve the reaction of boron trichloride (BCl₃) with hydrogen (H₂) in the presence of carbon at high temperatures (1700-1800 °C), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**: The reactants in the reaction are boron trichloride (BCl₃), hydrogen (H₂), and carbon (C).

2. **Understand the Reaction Conditions**: The reaction occurs at a high temperature (1700-1800 °C). High temperatures can lead to the formation of complex compounds.

3. **Predict the Products**: The reaction between BCl₃ and H₂ in the presence of carbon typically leads to the formation of boron carbide (B₄C) and hydrochloric acid (HCl).

4. **Write the Balanced Chemical Equation**:
- The unbalanced equation can be written as:
\[ BCl_3 + H_2 + C \rightarrow B_4C + HCl \]
- To balance the equation, we need to ensure that the number of each type of atom on the reactant side equals the number on the product side.
- Balancing the equation:
- There are 4 boron atoms in B₄C, so we need 4 BCl₃.
- Each BCl₃ has 3 Cl, so 4 BCl₃ gives 12 Cl, which means we need 12 HCl.
- To balance hydrogen, we need 6 H₂ (since each H₂ provides 2 H).
- Carbon remains as 1 in the product.
- The balanced equation is:
\[ 4 BCl_3 + 6 H_2 + C \rightarrow B_4C + 12 HCl \]

5. **Identify the Final Products**: From the balanced equation, the products formed are:
- Boron carbide (B₄C)
- Hydrochloric acid (HCl)

### Conclusion:
The products of the reaction between BCl₃, H₂, and C at high temperatures (1700-1800 °C) are boron carbide (B₄C) and hydrochloric acid (HCl).