Question

How much volume of chlorine is required to form 11.2L of hcl at 273K and 1atm pressure?

Answer 1

use the formula pv=nrt to get answer

Answer 2

0.0111 L volume of chlorine is required to form 11.2L of hcl at 273K and 1atm pressure.

Explanation:

Formation of HCl will take place as follows.

            $H_{2} + Cl_{2} \rightarrow 2HCl$

The data is given as follows.

            V = 11.2 L,    T = 273 K,   and P = 1 atm

Hence, calculate the number of moles as follows.

                          PV = nRT

    $1 atm \times 11.2 L = n \times 0.0821 atm L/mol K \times 273 K$

                      n = 0.499 mol

Therefore, according to the reaction equation 2 moles of HCl is formed by 1 mole of chlorine. Hence, volume of chlorine formed by 0.499 mol will be as follows.

                         $\frac{0.499 mol}{2}$

                           = 0.2495 mol

                           = $0.2495 mol \times 22.4 L/mol$

                           = 0.0111 L

Therefore, volume of chlorine will be 0.0111 liter.

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