標題:
chemistry question
發問:
In an experiment, 21.1 mL of 0.270 M barium hydroxide was mixed with 25.0 mL of 0.330 M aluminum sulfate. Based upon this reaction, you will be determiningWrite the complete ionic reaction that describes the reaction, including states, between barium hydroxide and aluminum sulfate.The concentration of the sulfate... 顯示更多 In an experiment, 21.1 mL of 0.270 M barium hydroxide was mixed with 25.0 mL of 0.330 M aluminum sulfate. Based upon this reaction, you will be determining Write the complete ionic reaction that describes the reaction, including states, between barium hydroxide and aluminum sulfate. The concentration of the sulfate anion is ?M and the concentration of the Al3+ cation is ?M.
最佳解答:
Equations: Al3+(aq) + 3OH-(aq) → Al(OH)3(s) Ba2+(aq) + SO42-(aq) → BaSO4(s) Originally, no. of moles of those ions are: Al3+(aq) = 2 × 0.33 × 0.025 = 0.0165 SO42-(aq) = 3 × 0.33 × 0.025 = 0.02475 Ba2+(aq) = 0.27 × 0.0211 = 0.005697 OH-(aq) = 0.27 × 0.0211 = 0.005697 So, for Ba2+ and SO42-, the latter is in excess with 0.02475 - 0.005697 = 0.019053 moles remained. For Al3+ and OH-, the former is in excess with 0.0165 - 0.005697/3 = 0.014601 moles remained. Now, resulting solution volume = 46.1 mL and therefore: [Al3+(aq)] = 0.014601/0.0461 = 0.317 M [SO42-(aq)] = 0.019053/0.0461 = 0.413 M
其他解答:
chemistry question
發問:
In an experiment, 21.1 mL of 0.270 M barium hydroxide was mixed with 25.0 mL of 0.330 M aluminum sulfate. Based upon this reaction, you will be determiningWrite the complete ionic reaction that describes the reaction, including states, between barium hydroxide and aluminum sulfate.The concentration of the sulfate... 顯示更多 In an experiment, 21.1 mL of 0.270 M barium hydroxide was mixed with 25.0 mL of 0.330 M aluminum sulfate. Based upon this reaction, you will be determining Write the complete ionic reaction that describes the reaction, including states, between barium hydroxide and aluminum sulfate. The concentration of the sulfate anion is ?M and the concentration of the Al3+ cation is ?M.
最佳解答:
Equations: Al3+(aq) + 3OH-(aq) → Al(OH)3(s) Ba2+(aq) + SO42-(aq) → BaSO4(s) Originally, no. of moles of those ions are: Al3+(aq) = 2 × 0.33 × 0.025 = 0.0165 SO42-(aq) = 3 × 0.33 × 0.025 = 0.02475 Ba2+(aq) = 0.27 × 0.0211 = 0.005697 OH-(aq) = 0.27 × 0.0211 = 0.005697 So, for Ba2+ and SO42-, the latter is in excess with 0.02475 - 0.005697 = 0.019053 moles remained. For Al3+ and OH-, the former is in excess with 0.0165 - 0.005697/3 = 0.014601 moles remained. Now, resulting solution volume = 46.1 mL and therefore: [Al3+(aq)] = 0.014601/0.0461 = 0.317 M [SO42-(aq)] = 0.019053/0.0461 = 0.413 M
其他解答:
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