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 balanced chemical equation for the reaction between zinc (Zn) and hydrochloric acid (HCl) is:
\[ \text{Zn}_{(s)} + 2 \text{HCl}_{(aq)} \rightarrow \text{ZnCl}_{2(aq)} + \text{H}_{2(g)} \]

### Step 2: Identify the initial 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 ratios
From the balanced equation, we see that:
- 1 mole of Zn reacts with 2 moles of HCl.

### Step 4: Calculate the amount of HCl needed for the available Zn
Since 1 mole of Zn requires 2 moles of HCl, the amount of HCl needed for 0.30 moles of Zn is:
\[ \text{HCl needed} = 0.30 \, \text{moles Zn} \times 2 \, \frac{\text{moles HCl}}{\text{mole Zn}} = 0.60 \, \text{moles HCl} \]

### Step 5: Identify the limiting reagent
We have only 0.52 moles of HCl available, which is less than the 0.60 moles required. Therefore, HCl is the limiting reagent.

### Step 6: Calculate the moles of H2 produced
From the balanced equation, the stoichiometric ratio of HCl to H2 is 2:1. Therefore, the moles of H2 produced from the available HCl can be calculated as follows:
\[ \text{Moles of H2} = \frac{\text{Moles of HCl}}{2} = \frac{0.52 \, \text{moles HCl}}{2} = 0.26 \, \text{moles H2} \]

### Conclusion
The number of moles of H2 produced is **0.26 moles**.

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