Question

0.0859g of compound X which contains a group II metal dissolves in water to produce colourless solution. Excess acidified silver nitrate was the added and 0.308g of a white precipitate (Y) was formed.
A) identify y
B) write an ionic equation for the formation of y
C) use the data provided to identify the group 2 metal present in compound X

Answer 1

Answer on Question#37737-Chemistry-Other

Question

0.0859 g of compound X, which contains a group II metal dissolves in water to produce colourless solution. Excess acidified silver nitrate was then added, and 0.308 g of a white precipitate (Y) was formed.

A) identify Y

B) write an ionic equation for the formation of Y

C) use the data provided to identify the group 2 metal present in compound X

Answer

A) $Y - \mathrm{AgCl}$ (silver chloride)

B) $\mathrm{Ag}^{+}_{\mathrm{(aq)}} + \mathrm{Cl}^{-}_{\mathrm{(aq)}} \rightarrow \mathrm{AgCl}_{\mathrm{(s)}}$

C) Let's design the group II metal present in compound X by M, than the X is $\mathrm{MCl}_2$, and the reaction:

$$\mathrm{MCl}_2 + 2 \mathrm{AgNO}_3 \rightarrow 2 \mathrm{AgCl} + \mathrm{M}(\mathrm{NO}_3)_2$$

Based on the reaction stoichiometry we have the proportion:

$$2 \cdot \mathrm{M}(\mathrm{AgCl}) - \mathrm{m}(\mathrm{AgCl})$$ $$\mathrm{M}(\mathrm{MCl}_2) - \mathrm{m}(\mathrm{MCl}_3)$$

Substituting the values we have:

$$2 \cdot 143.32 \ \mathrm{g/mol} - 0.308 \ \mathrm{g}$$ $$\mathrm{M}(\mathrm{MCl}_2) \ \mathrm{g/mol} - 0.0859 \ \mathrm{g}$$

Hence:

$$\mathrm{M}(\mathrm{MCl}_2) = 2 \cdot 143.32 \cdot 0.0859 / 0.308 = 79.94 \ \mathrm{g/mol}$$ $$\mathrm{M}(\mathrm{M}) = \mathrm{M}(\mathrm{MCl}_2) - 2 \cdot \mathrm{M}(\mathrm{Cl}) = 79.94 - 2 \cdot 35.45 = 9.04 \ \mathrm{g/mol}$$

So, the group II metal is **Beryllium**.