How many grams of al were reacted with excess hcl if 2.34 l of hydrogen gas were collected at stp in the following reaction2al + 6hcl --> 2alcl3 + 3h2

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