question 1
select all the product combinations that will react in a double displacement reaction
□ nacl and kno₃
□ hcl and na₂c₂o₄
□ hcl and na₂so₄
□ agno₃ and nacl

Question

Accepted Answer

# Explanation:
## Step 1: Recall Double Displacement Reaction Conditions
A double displacement reaction (metathesis) occurs if one of the products is a precipitate, a gas, or a weak electrolyte (like water). The general form is $AB + CD
ightarrow AD + CB$. We analyze each pair:

## Step 2: Analyze $NaCl$ and $KNO_3$
Possible products: $NaNO_3$ and $KCl$. Both are soluble salts (no precipitate, gas, or weak electrolyte). So, no reaction.

## Step 3: Analyze $HCl$ and $Na_2C_2O_4$
Reaction: $2HCl + Na_2C_2O_4
ightarrow 2NaCl + H_2C_2O_4$? Wait, no—actually, $H_2C_2O_4$ (oxalic acid) is a weak acid, but more importantly, the reaction produces $CO_2$ gas? Wait, no, let's check solubility. Wait, $HCl$ (strong acid) and $Na_2C_2O_4$ (sodium oxalate) react to form $NaCl$ and $H_2C_2O_4$? No, actually, $H_2C_2O_4$ can decompose, but more accurately, the reaction between a strong acid and a metal oxalate can produce $CO_2$ gas. Wait, the correct approach: check if products are precipitate, gas, or weak electrolyte. For $HCl + Na_2C_2O_4$, the products are $NaCl$ (soluble) and $H_2C_2O_4$ (weak acid, but also, $H_2C_2O_4$ can react further? Wait, no—actually, the key is that a double displacement reaction occurs if there's a driving force. For $HCl$ and $Na_2C_2O_4$, the reaction is $2HCl + Na_2C_2O_4
ightarrow 2NaCl + H_2C_2O_4$, but $H_2C_2O_4$ is a weak acid, but also, $H_2C_2O_4$ can decompose into $CO_2$, $CO$, and $H_2O$? No, that's decomposition. Wait, maybe I made a mistake. Let's re-express: the reaction between $HCl$ (acid) and $Na_2C_2O_4$ (salt of weak acid) produces a weak acid ($H_2C_2O_4$) and a salt ($NaCl$). Since $H_2C_2O_4$ is a weak electrolyte, this reaction occurs (acid-base or double displacement with a weak acid product).

## Step 4: Analyze $HCl$ and $Na_2SO_4$
Products: $NaCl$ and $H_2SO_4$. Both $NaCl$ (soluble) and $H_2SO_4$ (strong acid, soluble). No precipitate, gas, or weak electrolyte (except $H_2SO_4$ is strong). So, no reaction.

## Step 5: Analyze $AgNO_3$ and $NaCl$
Reaction: $AgNO_3 + NaCl
ightarrow AgCl(s) + NaNO_3$. $AgCl$ is a precipitate (insoluble in water). So, this reaction occurs (precipitate formation).

## Step 6: Recheck $HCl$ and $Na_2C_2O_4$
Wait, maybe the correct analysis: $HCl$ (aq) and $Na_2C_2O_4$ (aq) react to form $NaCl$ (aq) and $H_2C_2O_4$ (aq). But $H_2C_2O_4$ is a weak acid, so the reaction proceeds because a weak electrolyte is formed. Alternatively, maybe $H_2C_2O_4$ decomposes to $CO_2$, $CO$, and $H_2O$, but that's a subsequent reaction. However, the initial double displacement produces $H_2C_2O_4$, which is a weak electrolyte, so the reaction occurs. Wait, but let's confirm standard reactions:

- $NaCl + KNO_3$: no reaction (all soluble).
- $HCl + Na_2C_2O_4$: reaction (produces weak acid or gas? Wait, actually, $H_2C_2O_4$ is a weak acid, so the reaction occurs as a double displacement (acid + salt → new acid + new salt), where the new acid is weak.
- $HCl + Na_2SO_4$: no reaction (products are strong acid and soluble salt).
- $AgNO_3 + NaCl$: reaction (precipitate $AgCl$).

Wait, but some sources say that for double displacement, the driving force is precipitate, gas, or water. So $H_2C_2O_4$ is not a gas or water, but a weak electrolyte. Is that enough? Alternatively, maybe $HCl$ and $Na_2C_2O_4$ react to produce $CO_2$ gas. Let's write the reaction: $2HCl + Na_2C_2O_4
ightarrow 2NaCl + H_2C_2O_4$, and $H_2C_2O_4
ightarrow H_2O + CO_2 + CO$ (decomposition). But the initial double displacement produces $H_2C_2O_4$, which then decomposes. So the overall reaction has a gas, so the double displacement (the first step) is driven by the subsequent decomposition. So this reaction occurs.

So the reacting pairs are $HCl$ and $Na_2C_2O_4$, and $AgNO_3$ and $NaCl$.

# Answer:
The product combinations that react in a double displacement reaction are:
- $ \boldsymbol{HCl} $ and $ \boldsymbol{Na_2C_2O_4} $ (reacts to form a weak acid or gas)
- $ \boldsymbol{AgNO_3} $ and $ \boldsymbol{NaCl} $ (forms precipitate $ AgCl $)

(So the checkboxes to select are for $ \boldsymbol{HCl} $ and $ \boldsymbol{Na_2C_2O_4} $, and $ \boldsymbol{AgNO_3} $ and $ \boldsymbol{NaCl} $)

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QUESTION IMAGE

Question

Explanation:

Step 1: Recall Double Displacement Reaction Conditions

Step 2: Analyze \(NaCl\) and \(KNO_3\)

Step 3: Analyze \(HCl\) and \(Na_2C_2O_4\)

Step 4: Analyze \(HCl\) and \(Na_2SO_4\)

Step 5: Analyze \(AgNO_3\) and \(NaCl\)

Step 6: Recheck \(HCl\) and \(Na_2C_2O_4\)

Answer: