AgNO3(aq) + LiOH(aq) ---> AgOH(s) + LiNO3(aq)
更新1:
part 2: What would the percent yield be if 14.8 g of AgOH were made?
Chem Stoichiometry ?
2020-08-24 3:20 pm
How many grams of AgOH (molar mass = 169.9 g/mol) would be produced from 21.25 grams of AgNO3 (molar mass = 124.9 g/mol) according to the reaction below:
AgNO3(aq) + LiOH(aq) ---> AgOH(s) + LiNO3(aq)
AgNO3(aq) + LiOH(aq) ---> AgOH(s) + LiNO3(aq)
回答 (1)
2020-08-24 3:46 pm
AgNO₃(aq) + LiOH(aq) → AgOH(s) + LiNO₃(aq)
Mole ratio AgNO₃ : AgOH → 1 : 1
Moles of AgNO₃ reacted = (21.25 g) / (124.9 g/mol) = 0.17014 mol
Moles of AgOH produced = 0.17014 mol
Mass of AgOH produced = (0.17014 mol) × (169.9 g/mol) = 28.91 g
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OR:
(21.25 g AgNO₃) × (1 mol AgNO₃ / 124.9 g AgNO₃) × ( 1 mol AgOH / 1 mol AgNO₃) × (169.9 g AgOH / 1 mol AgOH)
= 28.91 g AgOH
Mole ratio AgNO₃ : AgOH → 1 : 1
Moles of AgNO₃ reacted = (21.25 g) / (124.9 g/mol) = 0.17014 mol
Moles of AgOH produced = 0.17014 mol
Mass of AgOH produced = (0.17014 mol) × (169.9 g/mol) = 28.91 g
====
OR:
(21.25 g AgNO₃) × (1 mol AgNO₃ / 124.9 g AgNO₃) × ( 1 mol AgOH / 1 mol AgNO₃) × (169.9 g AgOH / 1 mol AgOH)
= 28.91 g AgOH
收錄日期: 2021-04-24 07:57:51
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