Answer:
To determine the mass of the products when a large amount of MnS reacts with a small amount of HCl, you need to find the limiting reactant first.
1. Write the balanced chemical equation:
\[ \text{MnS(s) + 2HCl(aq) → MnCl2(aq) + H2S(g)} \]
2. Calculate the moles of each reactant using their molar masses:
\[ \text{Moles of MnS} = \frac{\text{mass of MnS}}{\text{molar mass of MnS}} \]
\[ \text{Moles of HCl} = \frac{\text{mass of HCl}}{\text{molar mass of HCl}} \]
3. Determine the limiting reactant by comparing the moles of each reactant and their stoichiometric coefficients in the balanced equation.
4. Once you know the limiting reactant, use it to calculate the moles of products formed.
5. Convert the moles of products to grams using their respective molar masses.
Without specific values for the molar masses and the actual masses of MnS and HCl, I can guide you through the steps but cannot provide the numerical result. If you provide the molar masses and the masses of MnS and HCl, I can assist further.
