Zinc reacts with 18.0 mL of 6.00 M cold, aqueous
sulfuric acid through single replacement.
a. How much zinc sulfate is produced?
Answer in units of g
Assuming zinc is in excess.
18 mL of 6.00 M H2SO4 equals 0.108 mol
From
Zn+H2SO4 -> ZnSO4 + H2
we deduce that 1 mol of H2SO4 produces 1 mol of ZnSO4.
So mass of ZnSO4 produced is 0.108 mol, multiplied by the molecular mass of ZnSO4.
To find the amount of zinc sulfate produced in grams, we need to first determine the limiting reagent, which is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
Step 1: Write and balance the chemical equation:
Zn + H2SO4 -> ZnSO4 + H2
Step 2: Calculate the moles of zinc (Zn) and sulfuric acid (H2SO4) used:
Moles of Zn = Molarity of Zn * Volume of Zn * Liters to milliliters conversion factor
Moles of H2SO4 = Molarity of H2SO4 * Volume of H2SO4 * Liters to milliliters conversion factor
Given:
Volume of H2SO4 = 18.0 mL
Molarity of H2SO4 = 6.00 M
Molarity of Zn = concentration of solid zinc (usually 1 M)
Step 3: Compare the moles of Zn and H2SO4 to determine the limiting reagent:
The balanced equation tells us that the stoichiometric ratio between Zn and H2SO4 is 1:1. Therefore, the mole ratio between them is the same.
Step 4: Calculate the moles of zinc sulfate (ZnSO4) produced:
Moles of ZnSO4 = Moles of limiting reagent (either Zn or H2SO4)
Step 5: Convert moles of ZnSO4 to grams using the molar mass of ZnSO4:
Molar mass of ZnSO4 = Molar mass of Zn + Molar mass of S + 4 * Molar mass of O
Finally, plug in the values and calculate the amount of zinc sulfate produced in grams.