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.