How many milliliters of a 0.5 molar HCl
solution are needed to react completely with
11 grams of zinc according to the following
equation?
? HCl + ? Zn ! ? ZnCl2 + ? H2
Answer in units of mL
See previous problem.
To determine the volume of the 0.5 molar HCl solution needed to react completely with 11 grams of zinc, we need to use stoichiometry and the equation balanced for the reaction.
First, we need to convert the mass of zinc (11 grams) to the number of moles. To do this, we use the molar mass of zinc (65.38 g/mol):
Number of moles of Zn = Mass of Zn / Molar mass of Zn
= 11 g / 65.38 g/mol
≈ 0.168 mol
Next, we need to find the molar ratio between HCl and Zn in the balanced equation. From the equation:
? HCl + ? Zn → ? ZnCl2 + ? H2
we can see that the stoichiometric coefficient for HCl is 2 because it reacts with 1 mole of Zn. Therefore, the molar ratio is 2 moles of HCl per 1 mole of Zn.
Now we can calculate the number of moles of HCl required to react with the given amount of zinc:
Number of moles of HCl = 2 * Number of moles of Zn
= 2 * 0.168 mol
≈ 0.336 mol
Finally, we can calculate the volume of HCl solution needed using the Molarity (0.5 M) and the formula:
Volume of HCl solution (in liters) = Number of moles of HCl / Molarity
= 0.336 mol / 0.5 mol/L
= 0.672 L
To convert the volume to milliliters, multiply the volume by 1000:
Volume of HCl solution (in mL) = 0.672 L * 1000
= 672 mL
Therefore, approximately 672 milliliters of the 0.5 molar HCl solution are needed to react completely with 11 grams of zinc.