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Chemistry question?

A student is given a sample of an unknown Group 2 chloride.
The student dissolves 2.86 g of the chloride in water.
The student adds excess aqueous silver nitrate.
8.604g of solid silver chloride, AgCl, forms.
(i) Calculate the amount, in moles, of AgCl that forms. The molar mass of AgCl = 143.4 g mol–1.

(ii) Deduce the amount, in moles, of the Group 2 chloride that the student dissolves. Hence deduce the relative atomic mass and the identity of the Group 2 metal.
Give the relative atomic mass to one decimal place. You must show your working.

This is the solution:


So the first bit is fairly easy, you are just using the equation of n=m/M= 8.604/143.4= 0.06 moles of AgCl

For the second part, it gets a little more tricker

Write a balanced equation of the reaction, as to work out moles, on the other 'reactants' side of the equation you can only write an equation, but you could also realise that as it is Cl2, the number of moles will be half, as it is Cl2:

XCl2 + 2AgNO3 = 2AgCl + XNO3

From the previous question we know the number of moles of AgCl is 0.06 moles and using the equation we can say for every 1 mole of XCl2 we produce 2 moles of AgCl, thus 0.06/2= 0.03 moles.

Now we can work the molar mass/Mr of XCl2= M=m/n= 2.86/0.03 which gives you 95.33, can you see where it has come from?

Now to work of the metal used, we do 91.33 -71 ( as its cl2) which should give you 24.3g, deducing that the metal used is Mg.

Hope this helped



I understand the solution;however i do not understand why the fact that the group 2 chloride may react with the water when dissolved and so we are ignoring the reaction that may take place here or am I just misunderstanding something?
Reply 1
Water is polar (H+ and O2-) so that means water that is not distilled will react with any ions present in the solution.
Original post by RandomStudent420
A student is given a sample of an unknown Group 2 chloride.
The student dissolves 2.86 g of the chloride in water.
The student adds excess aqueous silver nitrate.
8.604g of solid silver chloride, AgCl, forms.
(i) Calculate the amount, in moles, of AgCl that forms. The molar mass of AgCl = 143.4 g mol–1.

(ii) Deduce the amount, in moles, of the Group 2 chloride that the student dissolves. Hence deduce the relative atomic mass and the identity of the Group 2 metal.
Give the relative atomic mass to one decimal place. You must show your working.

This is the solution:


So the first bit is fairly easy, you are just using the equation of n=m/M= 8.604/143.4= 0.06 moles of AgCl

For the second part, it gets a little more tricker

Write a balanced equation of the reaction, as to work out moles, on the other 'reactants' side of the equation you can only write an equation, but you could also realise that as it is Cl2, the number of moles will be half, as it is Cl2:

XCl2 + 2AgNO3 = 2AgCl + XNO3

From the previous question we know the number of moles of AgCl is 0.06 moles and using the equation we can say for every 1 mole of XCl2 we produce 2 moles of AgCl, thus 0.06/2= 0.03 moles.

Now we can work the molar mass/Mr of XCl2= M=m/n= 2.86/0.03 which gives you 95.33, can you see where it has come from?

Now to work of the metal used, we do 91.33 -71 ( as its cl2) which should give you 24.3g, deducing that the metal used is Mg.

Hope this helped



I understand the solution;however i do not understand why the fact that the group 2 chloride may react with the water when dissolved and so we are ignoring the reaction that may take place here or am I just misunderstanding something?

MgCl2 doesn't react with water though

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