Problem 90
Question
Write the simplest chemical equation to represent the reaction of (a) \(\mathrm{K}_{2} \mathrm{CO}_{3}\left(\text { aq) and } \mathrm{Ba}(\mathrm{OH})_{2}(\mathrm{aq})\right.\); (b) \(\operatorname{Mg}\left(\mathrm{HCO}_{3}\right)_{2}(\) aq) on heating; (c) tin(II) oxide when heated with carbon; (d) \(\mathrm{CaF}_{2}(\mathrm{s})\) and \(\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{concd}\) aq); (e) \(\mathrm{NaHCO}_{3}(\mathrm{s})\) and \(\mathrm{HCl}(\mathrm{aq}) ;\) (f) \(\mathrm{PbO}_{2}(\mathrm{s})\) and HBr(aq); and (g) the reduction of \(\mathrm{SiF}_{4}\) to pure \(\mathrm{Si},\) by using Na as the reducing agent.
Step-by-Step Solution
Verified Answer
The reactions are: (a) \(K_2CO_3 + Ba(OH)_2 -> BaCO_3 + 2KOH\), (b) \(Mg(HCO_3)_2 -> MgCO_3 + CO_2 + H_2O\), (c) \(SnO + C -> Sn + CO\), (d) \(CaF_2 + H_2SO_4 -> CaSO_4 + 2HF\), (e) \(NaHCO_3+ HCl -> NaCl + CO_2 + H_2O\), (f) \(PbO_2+ 4HBr -> PbBr_2 + H_2O + Br_2\), (g) \(SiF_4 + 4Na -> 4NaF + Si\).
1Step 1: Reaction of K2CO3(aq) and Ba(OH)2(aq)
K2CO3(aq) + Ba(OH)2(aq) -> BaCO3(s) + 2KOH(aq). This is a double-displacement reaction where the carbonate ion from potassium carbonate combines with barium ion to form barium carbonate, which precipitates out of the solution.
2Step 2: Reaction of Mg(HCO3)2(aq) on heating
\(\mathrm{Mg(HCO_3)_2(aq)}\) -> \(\mathrm{MgCO_3 (s) + CO_2(g) + H_2O(g)}\). On heating, magnesium bicarbonate decomposes to form magnesium carbonate, carbon dioxide and water.
3Step 3: Reaction of tin(II) oxide when heated with carbon
\(\mathrm{SnO(s) + C(s)}\) -> \(\mathrm{Sn(s) + CO(g)}\). In this redox reaction, carbon reduces tin(II) oxide to elemental tin and is itself oxidized to carbon monoxide.
4Step 4: Reaction of CaF2(s) and H2SO4(concd)(aq)
\(\mathrm{CaF_2(s) + H_2SO_4(concd)(aq)}\) -> \(\mathrm{CaSO_4(s) + 2HF (g)}\). This is a double-displacement reaction where the fluoride ion from calcium fluoride displaces the sulfate ion in sulfuric acid.
5Step 5: Reaction of NaHCO3(s) and HCl(aq)
\(\mathrm{NaHCO_3(s))+ HCl(aq)\) -> \(\mathrm{NaCl(aq) + CO_2(g) + H_2O(l)}\). Sodium bicarbonate reacts with hydrochloric acid to form sodium chloride, carbon dioxide, and water.
6Step 6: Reaction of PbO2 and HBr(aq)
\(\mathrm{PbO_2(s)}\)+ 4HBr(aq) -> \(\mathrm{PbBr_2(aq) + H_2O(l) + Br_2(l)}\). Lead(II) oxide reacts with hydrobromic acid to create lead(II) bromide, water, and bromine.
7Step 7: Reduction of SiF4 to Si using Na
\(\mathrm{SiF_4(g)}\)+ 4Na(s) -> \(\mathrm{4NaF(s)+Si(s)}\). Silicon tetrafluoride reacts with sodium to form sodium fluoride and silicon. Sodium serves as the reducing agent in this reaction.
Key Concepts
Double-displacement reactionsDecomposition reactionsRedox reactions
Double-displacement reactions
Double-displacement reactions, also known as metathesis reactions, are a fascinating type of chemical reaction where two compounds exchange ions to form two new compounds. Imagine this as a dance, where partners swap places for an exciting new combination.
In double-displacement reactions, typically the ions from the original compounds switch. For example, in the reaction of potassium carbonate (\(\mathrm{K}_{2} \mathrm{CO}_{3}\)) with (\(\mathrm{Ba(OH)}_{2}\)), the carbonate ion (\(\mathrm{CO_{3}^{2-}}\)) from the potassium carbonate changes partners to join with the barium (\(\mathrm{Ba^{2+}}\)) forming barium carbonate (\(\mathrm{BaCO_{3}}\)), a solid precipitate that separates out of the solution.
In double-displacement reactions, typically the ions from the original compounds switch. For example, in the reaction of potassium carbonate (\(\mathrm{K}_{2} \mathrm{CO}_{3}\)) with (\(\mathrm{Ba(OH)}_{2}\)), the carbonate ion (\(\mathrm{CO_{3}^{2-}}\)) from the potassium carbonate changes partners to join with the barium (\(\mathrm{Ba^{2+}}\)) forming barium carbonate (\(\mathrm{BaCO_{3}}\)), a solid precipitate that separates out of the solution.
- This type of reaction is characterized by the formation of a precipitate, a gas, or a weak electrolyte.
- During the reaction of sodium bicarbonate with hydrochloric acid, carbon dioxide gas and water are two common products, demonstrating the versatility of double-displacement reactions in chemistry.
- Another example is when calcium fluoride reacts with sulfuric acid where hydrogen fluoride gas is evolved.
Decomposition reactions
Decomposition reactions involve a single compound breaking down into two or more simpler substances. Just picture it like a process of breaking a big complex Lego set back into simpler, individual blocks. These reactions are quite important in the chemical world as they give rise to elements and simpler compounds.
For example, when magnesium bicarbonate (\(\mathrm{Mg(HCO_{3})_{2}(aq)}\)) is heated, it decomposes to form magnesium carbonate (\(\mathrm{MgCO_{3}(s)}\)), along with carbon dioxide (\(\mathrm{CO_{2}(g)}\)) and water vapor (\(\mathrm{H_{2}O(g)}\)). This common laboratory demonstration shows how heat adds the energy needed to break bonds, leading to decomposition.
For example, when magnesium bicarbonate (\(\mathrm{Mg(HCO_{3})_{2}(aq)}\)) is heated, it decomposes to form magnesium carbonate (\(\mathrm{MgCO_{3}(s)}\)), along with carbon dioxide (\(\mathrm{CO_{2}(g)}\)) and water vapor (\(\mathrm{H_{2}O(g)}\)). This common laboratory demonstration shows how heat adds the energy needed to break bonds, leading to decomposition.
- Decomposition reactions are triggered by heat, light, or chemical energy.
- They are vital in everyday processes such as the breakdown of food in digestion or even the breakdown of pollutants in catalysis.
- Think about baking soda (\(\mathrm{NaHCO_{3}}\)) decomposition while baking pastries, which produces carbon dioxide, making the dough rise.
Redox reactions
Redox reactions, short for reduction-oxidation reactions, are essential chemical processes involving the transfer of electrons between two substances. These reactions are what keeps the world of chemistry balanced; while one substance loses electrons, the other gains them.
For instance, in the reaction between tin(II) oxide (\(\mathrm{SnO(s)}\)) and carbon (\(\mathrm{C(s)}\)), tin(II) oxide is reduced because it gains electrons to form elemental tin (\(\mathrm{Sn(s)}\)), whereas carbon is oxidized, losing electrons to form carbon monoxide (\(\mathrm{CO(g)}\)).
For instance, in the reaction between tin(II) oxide (\(\mathrm{SnO(s)}\)) and carbon (\(\mathrm{C(s)}\)), tin(II) oxide is reduced because it gains electrons to form elemental tin (\(\mathrm{Sn(s)}\)), whereas carbon is oxidized, losing electrons to form carbon monoxide (\(\mathrm{CO(g)}\)).
- Reduction refers to the gain of electrons, while oxidation involves the loss of electrons. Remember it with the mnemonic OIL RIG—Oxidation Is Loss, Reduction Is Gain.
- An excellent example includes the reaction of silicon tetrafluoride (\(\mathrm{SiF_{4}}\)) with sodium where sodium acts as a robust reducing agent.
- Redox reactions are everywhere around us from the rusting of iron to the energy production in our cells.
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