balance the reaction, do not leave any fractions, dont leave anything blank, then answer the questions about the reaction
__1__ir₂(so₄)₃ +__2__mg →__3__ir +__4__mgso₄
this is an __5__ reaction. is this a redox reaction? (yes/no)__6__, because ir⁺³ is reduced to ir⁰ and mg⁰ is __7__ to mg⁺².
will the reaction happen as it is written? (yes/no)__8__, because ir is a jewelry metal and is very 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

Question

Accepted Answer

# Explanation:
## Step1: Balance the sulfate ions
In $Ir_2(SO_4)_3$, there are 3 sulfate ($SO_4^{2 -}$) groups. In $MgSO_4$, there is 1 sulfate group. To balance the sulfate ions, the coefficient of $MgSO_4$ should be 3. So, we have $Ir_2(SO_4)_3+Mg
ightarrow Ir + 3MgSO_4$.
## Step2: Balance the magnesium atoms
Since there are 3 $Mg$ atoms in $3MgSO_4$, the coefficient of $Mg$ should be 3. Now the equation is $Ir_2(SO_4)_3 + 3Mg
ightarrow Ir+3MgSO_4$.
## Step3: Balance the iridium atoms
In $Ir_2(SO_4)_3$, there are 2 $Ir$ atoms. So the coefficient of $Ir$ should be 2. The balanced equation is $Ir_2(SO_4)_3+3Mg = 2Ir + 3MgSO_4$.

For the reaction type:
- In a single - displacement reaction, one element displaces another element in a compound. Here $Mg$ displaces $Ir$ from $Ir_2(SO_4)_3$, so it is a single - displacement reaction (V).
- A redox reaction is a reaction in which there is a transfer of electrons (change in oxidation states). $Ir^{3+}$ is reduced to $Ir^0$ (gain of electrons) and $Mg^0$ is oxidized (loss of electrons) to $Mg^{2+}$, so it is a redox reaction (Z).
- Since $Ir$ is a very stable jewelry metal, the reaction $Ir_2(SO_4)_3+3Mg = 2Ir + 3MgSO_4$ will not happen as written (AA) because $Mg$ is more reactive than $Ir$ and the reverse reaction (if $Ir$ were to displace $Mg$) is not favorable due to $Ir$'s stability.

# Answer:
1. A. 1
2. C. 3
3. A. 2
4. C. 3
5. V. single displacement
6. Z. yes
7. DD. oxidized
8. AA. no

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QUESTION IMAGE

Question

Explanation:

Step1: Balance the sulfate ions

Step2: Balance the magnesium atoms

Step3: Balance the iridium atoms

Answer: