4. What mass of aluminum hydroxide are needed to decompose in order to produce 65.0 L of water at STP according to the following equation? 2Al(OH)_(3)arrow 1Al_(2)O_(3)+3H_(2)O

## Step 1The balanced chemical equation given is:### \(2Al(OH)_{3} \rightarrow 1Al_{2}O_{3} + 3H_{2}O\)This equation tells us that 2 moles of Aluminum Hydroxide (Al(OH)3) decompose to produce 3 moles of water (H2O).## Step 2We are given that 65.0 L of water is produced. At Standard Temperature and Pressure (STP), 1 mole of any gas occupies 22.4 L. Therefore, we can calculate the number of moles of water produced by dividing the volume of water by the volume occupied by 1 mole of gas at STP.### ## Step 3From the balanced chemical equation, we 3 mo are produced from 2 moles of Aluminum Hydroxide. Therefore, we can calculate the number of moles of Aluminum Hydroxide needed by dividing the number of moles of water by 3.### \(Number\, of\, moles\, of\, Al(OH)_{3} = \frac{2.9\, moles}{3} = 0.97\, moles\)## Step 4The molar mass of Aluminum Hydroxide (Al(OH)3) is approximately 78 g/mol. Therefore, we can calculate the mass of Aluminum Hydroxide needed by multiplying the number of moles by the molar mass.### \(Mass\, of\, Al(OH)_{3} = 0.97\, moles \times 78\, g/mol = 75.5\, g\)