how many mililiters of a 0.27 M HCl solution are needed to react completely with 2.7 g of zinc to form zinc (II) chloride?"
equation.
2 HCl + 1 Zn -----> 1 ZnCl2 + 1 H2
To find out how many milliliters of a 0.27 M HCl (hydrochloric acid) solution are needed to react completely with 2.7 g of zinc, we need to use stoichiometry and the molarity (M) of the HCl solution.
First, let's find the number of moles of zinc (Zn) using its molar mass. The molar mass of Zn is 65.38 g/mol.
Number of moles of Zn = mass of Zn / molar mass of Zn
= 2.7 g / 65.38 g/mol
= 0.0413 mol
According to the balanced equation, the stoichiometry between HCl and Zn is 2:1. This means that 2 moles of HCl react with 1 mole of Zn.
Number of moles of HCl = 2 * number of moles of Zn
= 2 * 0.0413 mol
= 0.0826 mol
The molarity of the HCl solution is given as 0.27 M, which means that 0.27 moles of HCl are present in 1 liter (1000 mL) of solution.
Now, we can use the molarity to find the volume (in milliliters) of the HCl solution.
Volume of HCl solution = Number of moles of HCl / Molarity of HCl
= 0.0826 mol / 0.27 M
= 0.305 mL
Therefore, approximately 0.305 mL of the 0.27 M HCl solution is needed to react completely with 2.7 g of zinc to form zinc (II) chloride.
1. Convert 2.7 g Zn to moles. moles = grams/molar mass.
2. Using the coefficients in the balanced equation, convert moles Zn to moles HCl.
3. Now convert moles HCl to volume.
M = moles/L. You know M and moles, calculate L, then convert to mL.