Question

How many grams of KCl (s) has been produced from the thermal decomposition of KClO3 (s) that produced 50.0 mL of O2 (g) at 25 degrees C and 1.00 atm pressure.

Answer 1

Answer:

The mass is $m_{KCl} = 0.102 \ g$

Explanation:

From the question we are told that

   The volume of oxygen produced is  $V_o = 50mL = 50 *10^{-3} L$

    The temperature is $T = 25 ^oC = 25* 273 = 298 K$

     The pressure is  $P = 1.0\ atm$

From the ideal gas law we have that

      $PV = nRT$

Where R is the gas constant  with the value

         $R = 0.08206 \ atm \cdot L /mol K$

  n is the number of moles making it the subject of the formula

          $n = \frac{PV}{RT}$

Substituting values

          $n = \frac{1 * (50 *10^{-3}) }{0.08206 * 298}$

          $n = 2.045 *10^{-3} \ mol$

From the chemical equation

      one mole of  $KClO_3$ produces one mole of  kCl and   $\frac{3}{2}$ of oxygen

       x mole of  $KClO_3$ produces x mole of  kCl and   $2.045 *10^{-3} \ mol$ of oxygen

So    $x = \frac{2.045 *10^{-3}}{\frac{3}{2} }$

      $x = \frac{2}{3} * 2.045 *10^{-3}$

      $x = 1.363 *10^{-3} \ mol$

Now the molar mass of  KCl is a constant with a value

           $M_{KCl} = 74.55 g/mol$

Now the mass of KCl is mathematically evaluated as

          $m_{KCl} = x * M_{KCl}$

Substituting values

          $m_{KCl} = 1.363 *10^{-3} * 74.55$

         $m_{KCl} = 0.102 \ g$