Zinc reacts with hydrochloric acid (HCl(aq)) to form zinc chloride (ZnCl2) and hydrogen.
35.0 cm3 of 1.25M hydrochloric acid reacts with excess zinc, what is the percentage of yield if 1.80g of zinc chloride were formed?
Percentage yield?
2020-04-30 12:48 am
回答 (1)
2020-04-30 1:22 am
Molar mass of ZnCl₂ = (65.38 + 35.45×2) g/mol = 136.28 g/mol
Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)
Mole ratio HCl : ZnCl₂ = 2 : 1
Moles of HCl reacted = (1.25 mol/L) × (35.0/1000 L) = 0.04375 mol
Maximum moles of ZnCl₂ produced = (0.04375 mol) × (1/2) = 0.02188 mol
Theoretical yield of ZnCl₂ = (0.02188 mol) × (136.28 g/mol) = 2.98 g
Percentage of yield = (1.80/2.98) × 100% = 60.4%
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OR:
(1.25 mol HCl / 1000 mL HCl solution) × (35.0 mL HCl solution) × (1 mol ZnCl₂ / 2 mol HCl) × (136.28 g ZnCl₂ / 1 mol ZnCl₂)
= 2.98 g ZnCl₂ theoretical yield
(1.80/2.98) × 100% = 60.4% yield
Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)
Mole ratio HCl : ZnCl₂ = 2 : 1
Moles of HCl reacted = (1.25 mol/L) × (35.0/1000 L) = 0.04375 mol
Maximum moles of ZnCl₂ produced = (0.04375 mol) × (1/2) = 0.02188 mol
Theoretical yield of ZnCl₂ = (0.02188 mol) × (136.28 g/mol) = 2.98 g
Percentage of yield = (1.80/2.98) × 100% = 60.4%
====
OR:
(1.25 mol HCl / 1000 mL HCl solution) × (35.0 mL HCl solution) × (1 mol ZnCl₂ / 2 mol HCl) × (136.28 g ZnCl₂ / 1 mol ZnCl₂)
= 2.98 g ZnCl₂ theoretical yield
(1.80/2.98) × 100% = 60.4% yield
收錄日期: 2021-04-18 18:31:11
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20200429164812AA0SIv4