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Zn + 2 HCl -----> ZnCl2 + H2 1. How many liters of 12 M HCl are required to react with 14.0 grams of zinc?

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Final answer:

To react 14.0 grams of zinc with 12 M HCl, you would need 35.7 mL of the acid, based on stoichiometric calculations from the balanced reaction between zinc and hydrochloric acid.

Step-by-step explanation:

To determine how many liters of 12 M HCl are needed to react with 14.0 grams of zinc, we need to follow several steps:

  • First, we calculate the number of moles of zinc. The molar mass of zinc (Zn) is approximately 65.38 g/mol, so we divide the mass of zinc by its molar mass:
    14.0 g Zn / 65.38 g/mol = 0.214 mol Zn.
  • Using the balanced chemical equation Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g), we see that one mole of Zn reacts with two moles of HCl. This means we need 2 × 0.214 mol = 0.428 mol of HCl.
  • To find the volume of 12 M HCl needed, we use the following calculation:
    Volume = moles of HCl / concentration of HCl = 0.428 mol / 12 mol/L = 0.0357 L or 35.7 mL.

Therefore, you would need 35.7 mL of 12 M HCl to react completely with 14.0 grams of zinc.

User Jan Wilmans
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