Answer:
1.87 g of Al were reacted
Explanation:
We must apply the Ideal Gases law to solve the amount of moles of H₂ produced at STP
T = 273 K
P = 1 atm
P. V = n . R . T
1 atm . 2.34 L = n . 0.082 . 273 K
( 1 atm . 2.34 L ) / ( 0.082 . 273 K) = n
0.104 moles = n
Now we can find the moles of Al with the reaction. As ratio is 3:2 we have to make a rule of three to determine the moles we used.
3 moles of H₂ came from 2 moles of Al
0.104 moles of H₂ would come from ( 0.104 . 2 ) /3 = 0.0693 moles of Al
Finally we can know the mass of Al we used ( moles . molar mass)
0.0693 mol . 26.98 g/mol = 1.87 g
