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Answer :
The mass of zinc sulfate formed when 25 mL of 1.5 M sulfuric acid is added to excess zinc is approximately 6.05 grams.
To determine the mass of zinc sulfate (ZnSO4) formed when 25 mL of 1.5 M sulfuric acid is added to excess zinc, we need to consider the balanced chemical equation for the reaction between sulfuric acid and zinc:
H2SO4 + Zn -> ZnSO4 + H2
Given:
Volume of sulfuric acid (V) = 25 mL
Concentration of sulfuric acid (C) = 1.5 M
To calculate the moles of sulfuric acid, we can use the formula:
Moles = Volume (in liters) * Concentration (in moles per liter)
Moles of sulfuric acid = 25 mL * (1 L / 1000 mL) * 1.5 M
Next, we need to determine the stoichiometric ratio between sulfuric acid and zinc from the balanced equation, which is 1:1. This means that for every 1 mole of sulfuric acid, 1 mole of zinc sulfate is formed.
Since the problem states that there is an excess of zinc, all the sulfuric acid will react completely, forming an equal amount of zinc sulfate.
Therefore, the moles of zinc sulfate formed will be equal to the moles of sulfuric acid.
Now, we can calculate the mass of zinc sulfate using its molar mass. The molar mass of ZnSO4 is:
Zn: 65.38 g/mol
S: 32.07 g/mol
O: 16.00 g/mol (4 atoms)
Total molar mass of ZnSO4: 65.38 + 32.07 + 16.00(4) = 161.44 g/mol
Mass of zinc sulfate = Moles of ZnSO4 * Molar mass of ZnSO4
Mass of zinc sulfate = Moles of sulfuric acid * Molar mass of ZnSO4
Substituting the values:
Mass of zinc sulfate = (25 mL * (1 L / 1000 mL) * 1.5 M) * 161.44 g/mol
Calculating:
Mass of zinc sulfate ≈ 6.05 g
Therefore, the mass of zinc sulfate formed when 25 mL of 1.5 M sulfuric acid is added to excess zinc is approximately 6.05 grams.
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