Final answer:
To find the milliliters of 6.00 M HCl needed to react with 5.05 g of Zn, calculate the moles of Zn, use stoichiometry to find moles of HCl,
Step-by-step explanation:
The question asks how many milliliters of 6.00 M hydrochloric acid (HCl) would react with 5.05 g of zinc (Zn). The balanced chemical equation for the reaction between zinc and hydrochloric acid is Zn(s) + 2HCl(aq) \u2192 ZnCl₂(aq) + H₂(g).
To determine the volume of HCl needed, we must first calculate the moles of zinc reacted using its molar mass, then use the stoichiometry from the balanced equation to find the moles of HCl, and finally calculate the volume based on the molarity of HCl.
Zinc has a molar mass of 65.38 g/mol. Therefore, 5.05 g of Zn is:
5.05 g Zn / 65.38 g/mol = 0.0772 mol Zn
Since the reaction ratio from the balanced equation between Zn to HCl is 1:2, the moles of HCl required is 2 times the moles of Zn:
0.0772 mol Zn \u00d7 2 = 0.1544 mol HCl
To find the volume, use the molarity of the acid:
Volume HCl = Moles HCl / Molarity of HCl = 0.1544 mol / 6.00 M = 0.025733 L
Converting liters to milliliters:
0.025733 L \u00d7 1000 mL/L = 25.733 mL
Thus, 25.733 mL of 6.00 M HCl is required to completely react with 5.05 g of Zn.