2 K3PO4 + 3 MgCl2 → Mg3(PO4)2 + 6 KCl
Answer in units of mL.
What volume of 0.408 M K3PO4 is required to react with 28 mL of 0.75 M MgCl2 according to the equation?
2016-07-31 5:41 pm
回答 (3)
2016-07-31 5:53 pm
✔ 最佳答案
2 K₃PO₄ + 3 MgCl₂ → Mg₃(PO₄)₂ + 6 KClOR: Mole ratio K₃PO₄ : MgCl₂ = 2 : 3
No. of milli-moles of MgCl₂ = (0.75 mmol/mL) × (28 mL) = 21 mmol
No. of milli-moes of K₃PO₄ = (21 mmol) × (2/3) = 14 mmol
Volume of K₃PO₄ = (14 mmol) / (0.408 mmol/mL) = 34.3 mL
2016-07-31 5:51 pm
millimoles Mg(Cl)2 ---> (0.75 mmol/mL) (28 mL) = 21 mmol
K3PO4 and MgCl2 react in a 2:3 molar ratio
2 is to 3 as x is to 21
x = 14 mmol (of K3PO4 required)
0.408 mmol/mL = 14 mmol / y
y = 34.3 mL (round off to 34 if so desired)
K3PO4 and MgCl2 react in a 2:3 molar ratio
2 is to 3 as x is to 21
x = 14 mmol (of K3PO4 required)
0.408 mmol/mL = 14 mmol / y
y = 34.3 mL (round off to 34 if so desired)
2016-07-31 5:44 pm
no idea
收錄日期: 2021-04-18 15:23:14
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