Question

In the following chemical reaction between H​_2​ and Cl​_2​ to produce HCl, what is the mass of HCl produced and leftover reactants when 0.33 g of H​_2​ completely reacts with 12.46 g of Cl​_2​? H​_2​(g) + Cl​_2​(g) → 2HCl(g)

Answer 1

The balanced equation is

H2 + Cl2 ------> 2 HCl

Number of moles of H2 = 0.33 g / 2.016 g/mol = 0.164 mole

Number of moles of Cl2 = 12.46 g / 70.9060 g/mol = 0.176 mole

From the balanced equation we can say that

1 mole of H2 requires 1 mole of Cl2 so

0.164 mole of H2 will require

= 0.164 mole of H2 *(1 mole of Cl2 / 1 mole of H2)

= 0.164 mole of Cl2

But we have 0.176 mole of Cl2 which is in excess so Cl2 is an excess reactant and H2 is limiting reactant

From the balanced equation we can say that

1 mole of H2 produces 2 mole of HCl so

0.164 mole of H2 will produce

= 0.164 mole of H2 *(2 mole of HCl / 1 mole of H2)

= 0.328 mole of HCl

mass of 1 mole of HCl = 36.4609 g so

the mass of 0.328 mole of HCl = 12.0 g

Therefore, the mass of HCl produced would be 12.0 g

Number of moles of reactant left after completion of reaction = 0.176 - 0.164 = 0.0120 mole

mass of 1 mole of Cl2 = 70.9060 g so

the mass of 0.0120 mole of Cl2 = 0.851 g

Therefore, the mass of excess reactant left after completion of reaction = 0.851 g