Zinc and hydrochloric acid react according to the reaction :
`Zn_((s))+2HCl_((aq.))rarr ZnCl_(2(aq.))+H_(2(g))`
If `0.30` mole of `Zn` are added to hydrochloric acid containing `0.52` mole `HCl`, how many moles of `H_(2)` are produced ?

Options

A. `0.2`
B. `0.62`
C. `0.6`
D. `0.26`

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between zinc (Zn) and hydrochloric acid (HCl) can be represented by the balanced equation:
\[ \text{Zn}_{(s)} + 2\text{HCl}_{(aq)} \rightarrow \text{ZnCl}_{2(aq)} + \text{H}_{2(g)} \]

### Step 2: Identify the moles of reactants
From the problem, we have:
- Moles of Zn = 0.30 moles
- Moles of HCl = 0.52 moles

### Step 3: Determine the stoichiometric requirements
According to the balanced equation:
- 1 mole of Zn reacts with 2 moles of HCl to produce 1 mole of H2.

### Step 4: Calculate the amount of HCl needed for the available Zn
To find out how much HCl is required to react with 0.30 moles of Zn:
\[ \text{HCl required} = 0.30 \, \text{moles Zn} \times 2 \, \text{moles HCl/mole Zn} = 0.60 \, \text{moles HCl} \]

### Step 5: Identify the limiting reagent
We have only 0.52 moles of HCl available, but we need 0.60 moles of HCl to react with all the Zn. Therefore, HCl is the limiting reagent.

### Step 6: Calculate the moles of hydrogen gas produced
From the balanced equation, we see that:
- 2 moles of HCl produce 1 mole of H2.

Using the amount of HCl we have:
\[ \text{Moles of H2 produced} = \frac{0.52 \, \text{moles HCl}}{2} = 0.26 \, \text{moles H2} \]

### Conclusion
Thus, the number of moles of hydrogen gas (H2) produced is **0.26 moles**.

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