Question

balance the reaction, do not leave any fractions, dont leave anything blank, then answer the questions about the reaction
_1_au₂(so₄) +_2_al →_3_au + _4_al₂(so₄)₃
this is an _5_ reaction. is this a redox reaction? (yes/no)_6_, because au⁺¹ is _7_ to au⁰, and al⁰ is _8_ to al⁺³.
will the reaction happen as it is written? (yes/no)_9_, because au is a jewelry metal and is stable. the reverse reaction would not happen.
a. 1 b. 2 c. 3 d. 4 e. 5 f. 6 g. 7 h. 8
i. 9 j. 10 k. 12 l. 14 m. 16 n. 18 o. 20 p. 21
q. 24 r. 26 s. 28 t. synthesis u. decomposition
v. single displacement w. double displacement x. neutralization
y. combustion z. yes aa. no bb. reactive cc. stable
dd. oxidized ee. reduced

Explanation:

Step1: Balance sulfate ions

There are \(1\) sulfate ion in \(Au_2(SO_4)\) and \(3\) sulfate ions in \(Al_2(SO_4)_3\). To balance sulfate ions, we need to make the number of sulfate ions equal on both sides. So we put a coefficient of \(3\) in front of \(Au_2(SO_4)\).

$$3Au_2(SO_4)+Al ightarrow Au + Al_2(SO_4)_3$$

Step2: Balance gold atoms

Since we have \(3\times2 = 6\) gold atoms on the left - hand side (from \(3Au_2(SO_4)\)), we put a coefficient of \(6\) in front of \(Au\) on the right - hand side.

$$3Au_2(SO_4)+Al ightarrow6Au + Al_2(SO_4)_3$$

Step3: Balance aluminum atoms

There are \(2\) aluminum atoms in \(Al_2(SO_4)_3\). So we put a coefficient of \(2\) in front of \(Al\) on the left - hand side.

$$3Au_2(SO_4)+2Al ightarrow6Au + Al_2(SO_4)_3$$

For the reaction type:

• It is a single - displacement reaction (V), because one element (\(Al\)) displaces another element (\(Au\)) from its compound (\(Au_2(SO_4)\)).
• It is a redox reaction (Z). In \(Au^{+1}\) (in \(Au_2(SO_4)\)), the oxidation state of \(Au\) is \(+ 1\), and in \(Au^0\) (elemental \(Au\)), the oxidation state is \(0\). So \(Au^{+1}\) is reduced (EE) (gain of electrons, decrease in oxidation state). In \(Al^0\) (elemental \(Al\)), the oxidation state is \(0\), and in \(Al^{+3}\) (in \(Al_2(SO_4)_3\)), the oxidation state is \(+3\). So \(Al^0\) is oxidized (DD) (loss of electrons, increase in oxidation state).
• The reaction will not happen as written (AA). Gold (\(Au\)) is a very stable (CC) metal (a jewelry metal), and aluminum (\(Al\)) is more reactive than gold. But the standard reduction potential of \(Au^{+}\) is higher than that of \(Al^{3 +}\), and the reaction as written would require \(Al\) to reduce \(Au^{+}\) which is not favorable under normal conditions.

Answer:

1. \(3\)
2. \(2\)
3. \(6\)
4. \(1\)
5. V (single displacement)
6. Z (yes)
7. EE (reduced)
8. DD (oxidized)
9. AA (no)