✔ 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