how many milliliters of a 0.78 M HCl solution are needed to react completely with 6.4 g of zinc to form zinc(II) chloride? Answer in units of mL.
Here is a sample problem I posted on stoichiometry. You can use that. moles = M x L.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To find the volume of the HCl solution needed to react with the given amount of zinc, we can use the concept of molarity and the balanced chemical equation.
The balanced equation for the reaction between zinc and hydrochloric acid is:
Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g)
From the balanced equation, we can see that the stoichiometric ratio between zinc and hydrochloric acid is 1:2. This means that for every 1 mole of zinc, we need 2 moles of hydrochloric acid.
Let's calculate the number of moles of zinc using its molar mass:
Molar mass of zinc (Zn) = 65.38 g/mol
Number of moles of zinc = 6.4 g / 65.38 g/mol
Now, since we need 2 moles of hydrochloric acid for every 1 mole of zinc, we can convert the moles of zinc to moles of hydrochloric acid:
Number of moles of HCl = 2 moles of HCl / 1 mole of Zn × (number of moles of Zn)
Now, we can use the formula for molarity (M) to find the volume (in liters) of the hydrochloric acid solution needed:
Molarity (M) = moles of solute / volume of solution (in liters)
We rearrange the formula to solve for volume:
Volume of solution (in liters) = moles of solute / Molarity
Finally, we convert the volume from liters to milliliters:
Volume of solution (in mL) = Volume of solution (in liters) × 1000 mL
Let's plug in the values and calculate the answer:
Number of moles of Zn = 6.4 g / 65.38 g/mol ≈ 0.098 mol
Number of moles of HCl = 2 moles of HCl / 1 mole of Zn × (0.098 mol) ≈ 0.196 mol
Using the molarity (0.78 M) and the number of moles of HCl, we can find the volume of the HCl solution:
Volume of solution (in liters) = 0.196 mol / 0.78 M ≈ 0.251 L
Finally, converting the volume from liters to milliliters:
Volume of solution (in mL) = 0.251 L × 1000 mL ≈ 251 mL
Therefore, approximately 251 milliliters of a 0.78 M HCl solution are needed to react completely with 6.4 grams of zinc to form zinc(II) chloride.