Step-by-step analysis of each option:

Option A: $CH_3CH_2CH_2Cl \xrightarrow{Zn, dil. HCl}$

1. Identify the reaction type: This is the reduction of an alkyl halide (1-chloropropane) using a metal and acid ($Zn$ in dilute $HCl$).
2. Determine the reagents' function: The reaction of zinc with dilute HCl produces nascent hydrogen ($[H]$), which is a powerful reducing agent. The reaction is: $Zn + 2HCl \rightarrow ZnCl_2 + 2[H]$.
3. Write the chemical equation for the reduction: The alkyl halide is reduced to an alkane. The chlorine atom is replaced by a hydrogen atom. $CH_3CH_2CH_2Cl + 2[H] \xrightarrow{Zn, dil. HCl} CH_3CH_2CH_3 + HCl$
4. Identify the product: The product formed is propane ($CH_3CH_2CH_3$).
5. Conclusion: Since propane is the only organic product formed, it is the major product. Therefore, option A is correct.

Option B: $CH_3CH_2COONa \xrightarrow{NaOH, CaO, \Delta}$

1. Identify the reaction type: This is the soda-lime decarboxylation of a sodium salt of a carboxylic acid (sodium propanoate).
2. Determine the reagents' function: Soda-lime ($NaOH + CaO$) is a reagent used to remove the carboxylate group ($–COONa$) from the molecule.
3. Write the chemical equation: The reaction results in an alkane with one fewer carbon atom than the parent carboxylate salt. The $R-COONa$ group is converted to $R-H$. $CH_3CH_2COONa + NaOH \xrightarrow{CaO, \Delta} CH_3CH_3 + Na_2CO_3$
4. Identify the product: The starting material is sodium propanoate (3 carbons). The product is ethane ($CH_3CH_3$, 2 carbons).
5. Conclusion: The product is ethane, not propane. Therefore, option B is incorrect.

Option C: $CH_3CH_2Cl + CH_3Cl + Na \xrightarrow{dry \ ether}$

1. Identify the reaction type: This is a mixed Wurtz reaction involving two different alkyl halides (ethyl chloride and methyl chloride) with sodium metal in dry ether.
2. Predict the products: In a Wurtz reaction, alkyl radicals are formed which then couple. In a mixed reaction, three types of coupling can occur:
   • Self-coupling of ethyl chloride: $2 CH_3CH_2Cl + 2Na \rightarrow CH_3CH_2CH_2CH_3$ (n-butane)
   • Self-coupling of methyl chloride: $2 CH_3Cl + 2Na \rightarrow CH_3CH_3$ (ethane)
   • Cross-coupling of ethyl chloride and methyl chloride: $CH_3CH_2Cl + CH_3Cl + 2Na \rightarrow CH_3CH_2CH_3$ (propane)
3. Analyze the product distribution: The reaction produces a mixture of ethane, propane, and n-butane. To determine the major product, we consider the statistical probability of the radical couplings. If we assume equimolar amounts of reactants and equal reactivity, the statistical ratio of products (butane : propane : ethane) would be approximately 1:2:1. This is because there are two possibilities for cross-coupling ($CH_3\bullet + \bullet CH_2CH_3$) for every one possibility of self-coupling ($CH_3\bullet + \bullet CH_3$ or $CH_3CH_2\bullet + \bullet CH_2CH_3$).
4. Conclusion: Propane is formed in the highest yield among the alkane products. Thus, it can be considered a major product of this reaction mixture. Therefore, option C is correct.

Option D: $CH_3CH_2COONa(aq) \xrightarrow{electrolysis}$

1. Identify the reaction type: This is the Kolbe's electrolysis of an aqueous solution of a sodium salt of a carboxylic acid (sodium propanoate).
2. Predict the products: In Kolbe's electrolysis, the carboxylate ion is oxidized at the anode, loses $CO_2$, and forms an alkyl radical. These radicals then dimerize.
   • At the anode: $2 CH_3CH_2COO^- \rightarrow 2 CH_3CH_2\bullet + 2CO_2 + 2e^-$
   • Dimerization: $2 CH_3CH_2\bullet \rightarrow CH_3CH_2CH_2CH_3$ (n-butane)
3. Identify the major product: The main organic product from the coupling of ethyl radicals is n-butane. Side products from disproportionation (ethane and ethene) can also form but are generally minor. Propane is not a product of this reaction.
4. Conclusion: The major product is n-butane, not propane. Therefore, option D is incorrect.

Final Summary:

• Reaction A produces propane as the sole organic product.
• Reaction B produces ethane.
• Reaction C produces a mixture of alkanes where propane is the most abundant product.
• Reaction D produces n-butane as the major product.

Thus, the reactions that produce propane as a major product are A and C.