8.0 g of an element X react with an excess of copper (ll) tetraoxosulphate (lV) solution to deposit 21.3 g of copper. The correct equation for the reaction is?
(Cu = 64).
X(s) + CuSO4(aq) → Cu(s) + XSO4(aq)
X(s) + 2CuSO4(aq) → 2Cu(s) + X(SO4)2(aq)
2X(s) + CuSO4(aq) → Cu(s) + X2SO4(aq)
2X(s) + 3CuSO4(aq) → 3Cu(s) + X2(SO4)3(aq)
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8g of element X give 21.3g of copper then the amount of element X that will deposit 1 molecular mass of copper(64g) is x then x =8*64/21.3=24 and the element with mass number 24 is magnesium then the reaction is as follows Mg + CuSO4 = Cu + MgSO4 and it obey option A
This can be simply done by mere comparing the masses of the X elwment and that of the copper
8g of X element deposits 21.3g of copper
xg of X element deposits 64g of copper
Therefore x = 64×8/21.3
x = 24g
it means the atomic number of magnesium
Certainly, Toruka. Here's the full explanation written **in comment form**, suitable for use in an exam or write-up:
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**Comment:**
The reaction involves a metal X displacing copper from copper(II) tetraoxosulphate(VI), indicating a single displacement reaction. 21.3 g of Cu was deposited, which is equivalent to:
\[
\frac{21.3}{64} = 0.3328 \text{ mol of Cu}
\]
Since each Cu²⁺ ion needs 2 electrons to be reduced, the total electrons transferred is:
\[
0.3328 \times 2 = 0.6656 \text{ mol of electrons}
\]
The mass of X used is 8 g, so:
\[
\text{Moles of X} = \frac{8}{M}, \text{ where } M \text{ is the molar mass of X}
\]
If X donates 2 electrons per atom (as a divalent metal):
\[
\text{Moles of electrons from X} = 2 \times \frac{8}{M}
\]
Equating both:
\[
2 \times \frac{8}{M} = 0.6656 \Rightarrow M = \frac{16}{0.6656} \approx 24
\]
This molar mass corresponds to **magnesium**. Thus, X is magnesium, and the balanced chemical reaction is:
\[
\boxed{\text{Mg} + \text{CuSO}_4 \rightarrow \text{MgSO}_4 + \text{Cu}}
\]
This confirms that magnesium displaces copper, donating two electrons per atom.
