Answered

Join IDNLearn.com and become part of a knowledge-sharing community that thrives on curiosity. Discover in-depth and trustworthy answers to all your questions from our experienced community members.

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 )

Sagot :

Jufsanabriasaidn

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

We appreciate your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Accurate answers are just a click away at IDNLearn.com. Thanks for stopping by, and come back for more reliable solutions.