With the help of the balanced chemical equation
`2Al(s)+3H_(2)SO_(4)(aq) to Al_(2)(SO_(4))_(3)(aq)+3H_(2)(g)`
Answer the equations below :
How much Al (in grams) will be used to liberate 22.4 L of Hydrogen gas, at STP

Explanation

To solve the problem of how much aluminum (in grams) will be used to liberate 22.4 L of hydrogen gas at STP, we can follow these steps:

### Step 1: Understand the balanced chemical equation
The balanced chemical equation is:
\[ 2 \text{Al}(s) + 3 \text{H}_2\text{SO}_4(aq) \rightarrow \text{Al}_2(\text{SO}_4)_3(aq) + 3 \text{H}_2(g) \]

This equation tells us that 2 moles of aluminum react to produce 3 moles of hydrogen gas.

### Step 2: Determine the volume of hydrogen gas produced
At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 L. Since we are given that 22.4 L of hydrogen gas is produced, we can find the number of moles of hydrogen gas produced.

\[ \text{Moles of } H_2 = \frac{\text{Volume of } H_2}{\text{Volume of 1 mole at STP}} = \frac{22.4 \, \text{L}}{22.4 \, \text{L/mol}} = 1 \, \text{mol} \]

### Step 3: Relate moles of hydrogen to moles of aluminum
From the balanced equation, we see that 3 moles of hydrogen gas are produced from 2 moles of aluminum. We can set up a proportion to find out how many moles of aluminum are needed to produce 1 mole of hydrogen gas.

\[ \text{If } 3 \text{ moles of } H_2 \text{ require } 2 \text{ moles of Al, then } 1 \text{ mole of } H_2 \text{ requires } \frac{2}{3} \text{ moles of Al.} \]

### Step 4: Calculate the moles of aluminum needed
\[ \text{Moles of Al required} = \frac{2}{3} \text{ moles} \]

### Step 5: Convert moles of aluminum to grams
The molar mass of aluminum (Al) is approximately 27 g/mol. To find the mass of aluminum needed, we multiply the number of moles by the molar mass.

\[ \text{Mass of Al} = \text{Moles of Al} \times \text{Molar mass of Al} = \frac{2}{3} \text{ moles} \times 27 \, \text{g/mol} \]

Calculating this gives:
\[ \text{Mass of Al} = \frac{2 \times 27}{3} = \frac{54}{3} = 18 \, \text{g} \]

### Final Answer
Thus, the amount of aluminum required to liberate 22.4 L of hydrogen gas at STP is **18 grams**.

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