更新1:
I use systems of equations and get NaBr=.216 g. However, the answer in the book is .224 g. my answer does not check out but the book answer does. When I use NaBr=.216g, the masses add up to 2.06 g instead of 2 (verified using mass ratios).
A mixture of NaCl and NaBr have a mass of 2.0 g and the mass of Na is .75 g. HOW do you find the mass of NaBr? The book answer is NaBr=.224g?
回答 (3)
2017-12-28 3:05 am
✔ 最佳答案
Let z be the mass (in grams) of NaBr to be found.Then (2.00 - z) g is the mass of NaCl.
z / (102.8938 g NaBr/mol) x (1 mol Na / 1 mol NaBr) = (0.00971876 z) mol Na in NaBr
(2.00 - z) / (58.4430 g NaCl/mol) x (1 mol Na / 1 mol NaCl) = (0.0342214 - 0.0171107 z) mol Na in NaCl
(0.00971876 z) + (0.0342214 - 0.0171107 z) = (0.0342214 - 0.00739194 z) mol Na total
Given:
(0.0342214 - 0.00739194 z) mol Na x (22.98977 g Na/mol) = 0.75 g Na
0.786742 - 0.169939 z = 0.75
Solve for z algebraically:
z = 0.216207 = 0.216 g NaBr
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Checking:
(0.216 g NaBr) / (102.8938 g NaBr/mol) x (1 mol Na / 1 mol NaBr) = 0.00209925 mol Na in NaBr
(2.0 - 0.216) g NaCl / (58.4430 g NaCl/mol) x (1 mol Na / 1 mol NaCl) = 0.0305255 mol Na in NaCl
(0.00209925 mol Na + 0.0305255 mol Na) x (22.98977 g Na/mol) = 0.750035 g Na
2017-12-28 1:02 am
Let y g be the mass of NaBr.
Then, mass of NaCl = (2.0 - y) g
Mass fraction of Na in NaBr = 22.99 / (22.99 + 79.90) = 22.99/102.89
Mass fraction of Na in NaCl = 22.99 / (22.99 + 35.45) = 22.99/58.44
Total mass of Na in the mixture = 0.75 g
(Mass of Na in NaBr) + (Mass of Na in NaCl) = 0.75 g
[y × (22.99/102.89)] + [(2.0 - y) × (22.99/58.44)] = 0.75
(22.99y/102.89) + (45.98/58.45) - (22.99y/58.44) = 0.75
(22.99y/102.89) - (22.99y/58.45) = 0.75 - (45.98/58.45)
22.99y × [(1/102.89) - (1/58.45)] = 0.75 - (45.98/58.45)
y = [0.75 - (45.98/58.45)] / {22.99 × [(1/102.89) - (1/58.45)]}
y = 0.216
Mass of NaBr in the mixture = 0.216 g
Then, mass of NaCl = (2.0 - y) g
Mass fraction of Na in NaBr = 22.99 / (22.99 + 79.90) = 22.99/102.89
Mass fraction of Na in NaCl = 22.99 / (22.99 + 35.45) = 22.99/58.44
Total mass of Na in the mixture = 0.75 g
(Mass of Na in NaBr) + (Mass of Na in NaCl) = 0.75 g
[y × (22.99/102.89)] + [(2.0 - y) × (22.99/58.44)] = 0.75
(22.99y/102.89) + (45.98/58.45) - (22.99y/58.44) = 0.75
(22.99y/102.89) - (22.99y/58.45) = 0.75 - (45.98/58.45)
22.99y × [(1/102.89) - (1/58.45)] = 0.75 - (45.98/58.45)
y = [0.75 - (45.98/58.45)] / {22.99 × [(1/102.89) - (1/58.45)]}
y = 0.216
Mass of NaBr in the mixture = 0.216 g
2017-12-27 9:30 pm
Molar mass NaCl = 58.55g/mol
Molar mass NaBr = 100.89g/mol
Molar mass Na = 22.99g/mol
Mass of Na in NaCl = Xg
Then mass Na in NaBr = (0.75 - X)g
Mass of NaCl = 58.55/22.99*( X) = 2.546Xg
Mass of Na in NaBr = 100.89 / 22.99 * ( 0.75-X)
Mass of Na in NaBr = 4.388 ( 0.75 - X)
Mass of Na in NaBr = 3.291 - 4.388X g
2.546Xg + 3.291g - 4.388Xg = 2g
2.546X - 4.388X = 2g - 3.291g
-1.842X = - 1.291
X = 0.700g
Mass of NaCl = 58.55/22.99*0.700 = 1.785g
Mass of NaBr = 2.00g - 1.785g = 0.215g
I agree with your answer.
Molar mass NaBr = 100.89g/mol
Molar mass Na = 22.99g/mol
Mass of Na in NaCl = Xg
Then mass Na in NaBr = (0.75 - X)g
Mass of NaCl = 58.55/22.99*( X) = 2.546Xg
Mass of Na in NaBr = 100.89 / 22.99 * ( 0.75-X)
Mass of Na in NaBr = 4.388 ( 0.75 - X)
Mass of Na in NaBr = 3.291 - 4.388X g
2.546Xg + 3.291g - 4.388Xg = 2g
2.546X - 4.388X = 2g - 3.291g
-1.842X = - 1.291
X = 0.700g
Mass of NaCl = 58.55/22.99*0.700 = 1.785g
Mass of NaBr = 2.00g - 1.785g = 0.215g
I agree with your answer.
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