Use the Conservation of Mass Interactive to answer the question.

Consider the reaction.

MnS (s) + 2HCl (aq) ⟶MnCl2 (aq) + H2S (g)

If the small amount of MnS is combined with the large amount of HCl, what is the mass of the products? The mass of any excess reactants will also be included in this total

The mass of the beaker is 400.00g. The mass of the beaker with the large amount of MnS (s) and HCl (aq) after reaction is 508.76g. The mass of the beaker with the small amount of MnS (s) and HCl (aq) after reaction is 454.35g.

The number of moles of product formed = 110.88g

Explanation:

MnS + 2HCl → Mn Cl_2 + h_2S

1 mole of MnS reacts with 2 moles of HCl to give 1 mole of MnCl_2 and 1 mole of H_2S.

If HCl is taken in large amount, then MnS is limiting reagent. Thus the formulation of the product depends upon the moles of MnS taken.

Let the moles of MnS = xg

Mass of MnS and HCl = 508.7 - 400 = 108.76 g

Moles = mass / molar mass

Moles of MnCl_2 = 108.7 / 123.5 = 0.880

The number of moles of product formed = moles of MnCl_2 * molar mass of MnCl_2

=0.880 * 126g

=110.88g

The number of moles of product formed = 110.88g