how many mililiters of a 0.4 molar HCl solution are needed to react completely with 13 grams of zinc according to the following equation?
Equation.
2 HCl + 1 Zn -----> 1 ZnCl2 + 1 H2
13 g of Zn is 13/65.4 = 0.199 moles. You might as well round that off to 0.20 moles. You will require 0.4 moles of HCl for complete reaction.
That would be contained in 1 liter of a 0.4 molar solution.
One liter is 1000 ml.
To find out how many milliliters of a 0.4 molar HCl solution are needed to react completely with 13 grams of zinc, we need to use the equation and consider the molar ratios.
First, we need to convert the mass of zinc (13 grams) to moles. To do this, we'll divide the mass by the molar mass of zinc. The molar mass of zinc (Zn) is approximately 65.38 grams/mole.
Moles of Zn = Mass of Zn / Molar mass of Zn
Moles of Zn = 13 grams / 65.38 grams/mole
Moles of Zn ≈ 0.199 moles
Now, let's determine the ratio between the moles of HCl and Zn from the balanced equation. According to the equation, it's a 2:1 ratio of moles of HCl to moles of Zn.
Moles of HCl = 2 * Moles of Zn
Moles of HCl = 2 * 0.199 moles
Moles of HCl ≈ 0.398 moles
Finally, to determine the volume of the HCl solution in milliliters, we'll use the molarity (0.4 M) and the moles of HCl.
Molarity (M) = Moles / Volume
Volume (in liters) = Moles / Molarity
Volume (in milliliters) = (Moles / Molarity) * 1000
Volume (in milliliters) = (0.398 moles / 0.4 moles/liter) * 1000
Volume (in milliliters) ≈ 995 milliliters
Therefore, approximately 995 milliliters of a 0.4 molar HCl solution are needed to react completely with 13 grams of zinc according to the given equation.