In order to determine the formula of a chloride of the element, Q, 0.1 mole of the chloride was dissolved in 500cm3 of water. 50cm3 of this solution just reacted with 300cm3 of 0.1M AgNO3(aq). Which one of the following is a possible formula for the chloride?
A. Q3Cl
B. Q2Cl6
C. QCl
D. Q2Cl3
E. QCl6
Can anyone help with this question?
2017-10-29 8:31 pm
回答 (1)
2017-10-29 9:31 pm
No. of moles of Ag⁺ from AgNO₃ = (0.1 mol/dm³) × (300/1000 dm³) = 0.03 mol
In AgCl, mole ratio Ag⁺ : Cl⁻ = 1 : 1
No. of moles of Ag⁺ in AgCl formed = 0.03 mol
No. of moles of Cl⁻ in AgCl formed = 0.03 mol
No. of moles of Cl⁻ in 50 cm³ of the solution = 0.03 mol
No. of moles of Cl⁻ in 500 cm³ of the solution = (0.03 mol) × (500/50) = 0.3 mol
No. of moles of Cl⁻ in 0.1 mol of the chloride of Q = 0.3 mol
Hence, the formula of the chloride is QnCl₃.
The answer: D. Q₂Cl₃
In AgCl, mole ratio Ag⁺ : Cl⁻ = 1 : 1
No. of moles of Ag⁺ in AgCl formed = 0.03 mol
No. of moles of Cl⁻ in AgCl formed = 0.03 mol
No. of moles of Cl⁻ in 50 cm³ of the solution = 0.03 mol
No. of moles of Cl⁻ in 500 cm³ of the solution = (0.03 mol) × (500/50) = 0.3 mol
No. of moles of Cl⁻ in 0.1 mol of the chloride of Q = 0.3 mol
Hence, the formula of the chloride is QnCl₃.
The answer: D. Q₂Cl₃
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