Calcule la masa en gramos de cloruro de hidrógeno (HCl) que se forma cuando 5.6 L de hidrógeno molecular, medido a TPE, reacciona con un exceso de cloro molecular gaseoso. MHCl = 36.5 g/mol H2(g) + Cl2(g) → 2HCl(g)

Calculate the mass in grams of hydrogen chloride (HCl) that is formed when 5.6 L of molecular hydrogen, measured at TPE, reacts with an excess of gaseous molecular chlorine. MHCl = 36.5 g/mol

H2(g) + Cl2(g) → 2HCl(g)

18.25g

Explanation:

The reaction equation is given as;

        H₂   +   Cl₂   →   2HCl

Given parameters:

Volume of hydrogen gas = 5.6L

 The reaction occurs at STP.

Solution:

To solve this problem, let us find the number of moles of the hydrogen as;

          At STP;

    22.4L of a gas has 1 mole;

    5.6L of Hydrogen as will have   = 0.25mole

From the balanced reaction equation;

        1 mole of H₂  will produce 2 mole of HCl

      0.25 mole of H₂ will produce 2 x 0.25 = 0.5mole of HCl

The mass of HCl';

    Mass of HCl  = number of moles x molar mass

   Mass of HCl  = 0.5 x 36.5  = 18.25g

it means there are 3 atoms of oxygen

the answer is b. dilution