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Answer :
Final answer:
The mass of Zn needed to react with 17.3 mL of 1.55 M HCl is approximately 0.876 grams.
Explanation:
To calculate the mass of Zn needed to react with 17.3 mL of 1.55 M HCl, we need to use the balanced chemical equation for the reaction between Zn and HCl:
Zn + 2HCl → ZnCl2 + H2
From the balanced equation, we can see that the molar ratio between Zn and HCl is 1:2. This means that for every 1 mole of Zn, we need 2 moles of HCl.
First, we need to calculate the number of moles of HCl using the given concentration and volume:
Number of moles of HCl = concentration × volume = 1.55 M × 0.0173 L = 0.026815 moles
Since the molar ratio between Zn and HCl is 1:2, we need half the number of moles of Zn:
Number of moles of Zn = 0.026815 moles ÷ 2 = 0.0134075 moles
Finally, we can calculate the mass of Zn using its molar mass:
Mass of Zn = number of moles × molar mass = 0.0134075 moles × 65.38 g/mol = 0.876 g
Learn more about calculating mass of a reactant here:
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