Answer :
Sure, let's balance and identify the products of each chemical reaction and write the full balanced equations.
a. [tex]\( Fe (s) + Pb \left( NO _3 \right)_2 ( aq ) \rightarrow \)[/tex]
This is a single displacement reaction where iron (Fe) will displace lead (Pb) in lead(II) nitrate. Iron is more reactive than lead, so iron will replace lead in the compound.
Balanced equation:
[tex]\[ Fe (s) + Pb \left( NO _3 \right)_2 ( aq ) \rightarrow Fe \left( NO _3 \right)_2 ( aq ) + Pb (s) \][/tex]
b. [tex]\( Cl _2 (aq) + 2 NaI (aq) \rightarrow \)[/tex]
This is a single displacement reaction where chlorine (Cl[tex]\(_2\)[/tex]) displaces iodine (I[tex]\(_2\)[/tex]) from sodium iodide (NaI). Chlorine is more reactive than iodine, so it will take iodine's place in the compound.
Balanced equation:
[tex]\[ Cl _2 (aq) + 2 NaI (aq) \rightarrow 2 NaCl (aq) + I _2 (s) \][/tex]
c. [tex]\( Ca (s) + 2 H _2 O (l) \rightarrow \)[/tex]
This is a single displacement reaction where calcium (Ca) reacts with water (H[tex]\(_2\)[/tex]O) to form calcium hydroxide (Ca(OH)[tex]\(_2\)[/tex]) and hydrogen gas (H[tex]\(_2\)[/tex]).
Balanced equation:
[tex]\[ Ca (s) + 2 H _2 O (l) \rightarrow Ca(OH)_2 (aq) + H _2 (g) \][/tex]
d. [tex]\( Br _2 (aq) + NaCl (aq) \rightarrow \)[/tex]
Bromine (Br[tex]\(_2\)[/tex]) is not reactive enough to displace chlorine (Cl[tex]\(_2\)[/tex]) from sodium chloride (NaCl). Since Br[tex]\(_2\)[/tex] and Cl[tex]\(_2\)[/tex] are both halogens and in the same group of the periodic table, reactivity differences are necessary to displace one by the other. Here, no reaction will occur.
Balanced equation:
[tex]\[ Br _2 (aq) + NaCl (aq) \rightarrow \text{No reaction} \][/tex]
To summarize, the balanced equations for the reactions are:
a. [tex]\[ Fe (s) + Pb \left( NO _3 \right)_2 ( aq ) \rightarrow Fe \left( NO _3 \right)_2 ( aq ) + Pb (s) \][/tex]
b. [tex]\[ Cl _2 (aq) + 2 NaI (aq) \rightarrow 2 NaCl (aq) + I _2 (s) \][/tex]
c. [tex]\[ Ca (s) + 2 H _2 O (l) \rightarrow Ca(OH)_2 (aq) + H _2 (g) \][/tex]
d. [tex]\[ Br _2 (aq) + NaCl (aq) \rightarrow \text{No reaction} \][/tex]
a. [tex]\( Fe (s) + Pb \left( NO _3 \right)_2 ( aq ) \rightarrow \)[/tex]
This is a single displacement reaction where iron (Fe) will displace lead (Pb) in lead(II) nitrate. Iron is more reactive than lead, so iron will replace lead in the compound.
Balanced equation:
[tex]\[ Fe (s) + Pb \left( NO _3 \right)_2 ( aq ) \rightarrow Fe \left( NO _3 \right)_2 ( aq ) + Pb (s) \][/tex]
b. [tex]\( Cl _2 (aq) + 2 NaI (aq) \rightarrow \)[/tex]
This is a single displacement reaction where chlorine (Cl[tex]\(_2\)[/tex]) displaces iodine (I[tex]\(_2\)[/tex]) from sodium iodide (NaI). Chlorine is more reactive than iodine, so it will take iodine's place in the compound.
Balanced equation:
[tex]\[ Cl _2 (aq) + 2 NaI (aq) \rightarrow 2 NaCl (aq) + I _2 (s) \][/tex]
c. [tex]\( Ca (s) + 2 H _2 O (l) \rightarrow \)[/tex]
This is a single displacement reaction where calcium (Ca) reacts with water (H[tex]\(_2\)[/tex]O) to form calcium hydroxide (Ca(OH)[tex]\(_2\)[/tex]) and hydrogen gas (H[tex]\(_2\)[/tex]).
Balanced equation:
[tex]\[ Ca (s) + 2 H _2 O (l) \rightarrow Ca(OH)_2 (aq) + H _2 (g) \][/tex]
d. [tex]\( Br _2 (aq) + NaCl (aq) \rightarrow \)[/tex]
Bromine (Br[tex]\(_2\)[/tex]) is not reactive enough to displace chlorine (Cl[tex]\(_2\)[/tex]) from sodium chloride (NaCl). Since Br[tex]\(_2\)[/tex] and Cl[tex]\(_2\)[/tex] are both halogens and in the same group of the periodic table, reactivity differences are necessary to displace one by the other. Here, no reaction will occur.
Balanced equation:
[tex]\[ Br _2 (aq) + NaCl (aq) \rightarrow \text{No reaction} \][/tex]
To summarize, the balanced equations for the reactions are:
a. [tex]\[ Fe (s) + Pb \left( NO _3 \right)_2 ( aq ) \rightarrow Fe \left( NO _3 \right)_2 ( aq ) + Pb (s) \][/tex]
b. [tex]\[ Cl _2 (aq) + 2 NaI (aq) \rightarrow 2 NaCl (aq) + I _2 (s) \][/tex]
c. [tex]\[ Ca (s) + 2 H _2 O (l) \rightarrow Ca(OH)_2 (aq) + H _2 (g) \][/tex]
d. [tex]\[ Br _2 (aq) + NaCl (aq) \rightarrow \text{No reaction} \][/tex]
