Answer:
82.9 L of H2
Step-by-step explanation:
We need to start with a balanced equation. The following looks correct:
2NaOH+2Al+2H2O = 2NaAlO2 + 3H2
This tells us to expect 3 moles of H2 gas for every 2 moles of sodium hydroxide. It also tells us that we need 2 moles of Al for every 2 moles of NaOH.
Now calculate the moles of both the sodium hydroxide and the aluminum:
NaOH: The molar mass is 40.0 g/mole
(75g NaOH)/(40.0 g/mole) = 1.875 moles NaOH
Al: The molar mass is 27.0 g/mole
(50.0g Al)/(27.0 g/mole) = 1.85 moles Al
The values are close, but the reaction will stop when all the moles of Al are consumed, leaving 0.025 moles of NaOH, since they are consumed at the same proportion, as per the balanced equation.
The equation says we should expect 3 moles H2 gas for evey 2 moles of Al. So:
(1.85 moles Al)*(3 moles H2)/(2 moles Al) = 3.7 moles of H2
At STP, the conversion factor for all gasses is 22.4L/mole (STP only).
(3.7 moles of H2)*(22.4L/mole) = 82.9 L of H2