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