Question

2Al + 6HCl --> 2AlCl3 + 3H2 Aluminium reacts with hydrochloric acid. How many grams of aluminum are necessary to produce 11 L of hydrogen gas at STP? ( 8.8 g )

Answer 1

Answer:

8.8g of Al are necessaries

Explanation:

Based on the reaction, 2 moles of Al are required to produce 3 moles of hydrogen gas.

To solve this question we must find the moles of H2 in 11L at STP using PV = nRT. With these moles we can find the moles of Al required and its mass as follows:

Moles H2:

PV = nRT; PV/RT = n

Where P is pressure = 1atm at STP; V is volume = 11L; R is gas constant = 0.082atmL/molK and T is absolute temperature = 273.15K at STP

Replacing:

1atm*11L/0.082atmL/molK*273.15K = n

n = 0.491 moles of H2 must be produced

Moles Al:

0.491 moles of H2 * (2mol Al / 3mol H2) = 0.327moles of Al are required

Mass Al -Molar mass: 26.98g/mol-:

0.327moles of Al * (26.98g / mol) = 8.8g of Al are necessaries