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How many grams of ZnS(s) are precipitated when 30.0 milliliters of 1.76

1 Answer

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Final answer:

To find the mass of ZnS precipitated, you would need to use the volume and molarity of the solution to calculate moles, then use the molar mass of ZnS (97.44 g/mol) to find the mass. However, the question lacks complete data to provide a precise answer.

Step-by-step explanation:

The question does not provide the complete information on what the 1.76 is related to (like the molarity of a solution which would be essential information to calculate the amount of ZnS precipitate). However, I can guide you on how to solve the problem if you had the concentration.

If 1.76 M is the concentration of a solution that reacts with Zn2+ ions to precipitate ZnS, you would first calculate the number of moles of reactant using the volume and molarity (M).

To find the mass of precipitated ZnS:

  1. Calculate moles of the reactant from volume (V) and molarity (M) using the formula:
  2. moles = M × V
  3. Use the stoichiometry from the balanced chemical reaction to find moles of ZnS precipitated.
  4. Calculate the mass of ZnS precipitated using the molar mass of ZnS (97.44 g/mol).

Note: To answer completely, the reaction and molarity of the other reactant are needed.

User Ameer Deen
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