Zn(s)+2HCl(aq)⟶ZnCl2(aq)+H2(g) How many milliliters of 2.00 M HCl(aq) are required to react with 4.85 g Zn(s)? volume:?
回答 (1)
2020-09-23 2:24 pm
Molar mass of Zn = 65.4 g/mol
Balance equation for the reaction:
Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)
Mole ratio Zn : HCl = 1 : 2
Moles of Zn reacted = (4.85 g) / (65.4 g/mol) = 0.07416 mol
Moles of HCl required = (0.07416 mol) × 2 = 0.1483 mol
Volume of HCl required = (0.1483 mol) / (2.00 mol/L) = 0.0742 L = 74.2 mL
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OR:
(4.85 g Zn) × (1 mol Zn / 65.4 g Zn) × (2 mol HCl / 1 mol HCl) × (1000 mL HCl / 2.00 mol HCl)
= 74.2 mL HCl
Balance equation for the reaction:
Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)
Mole ratio Zn : HCl = 1 : 2
Moles of Zn reacted = (4.85 g) / (65.4 g/mol) = 0.07416 mol
Moles of HCl required = (0.07416 mol) × 2 = 0.1483 mol
Volume of HCl required = (0.1483 mol) / (2.00 mol/L) = 0.0742 L = 74.2 mL
====
OR:
(4.85 g Zn) × (1 mol Zn / 65.4 g Zn) × (2 mol HCl / 1 mol HCl) × (1000 mL HCl / 2.00 mol HCl)
= 74.2 mL HCl
收錄日期: 2021-05-01 09:36:49
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