Question

According to the reaction represented above, about how many grams of aluminum (atomic mass 27 g) are necessary to produce 0. 50 mol of hydrogen gas at 25ºc and 1. 00 atm?.

Answer 1

Answer:

Taking into account the reaction stoichiometry, the correct answer is option B): 9 grams of aluminum are necessary to produce 0.50 mol of hydrogen gas at 25°C and 1.0 atm.

Explanation:

The balanced reaccion is:

6 HCl + 2 Al → 2 AlCl₃ + 3 H₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

HCl: 6 moles

Al: 2 moles

AlCl₃: 2 moles

H₂: 3 moles

The atomic mass of aluminum is 27 g/mole. Then, by reaction stoichiometry, the following mass quantity of aluminum participates in the reaction:

Then you can apply the following rule of three: if by reaction stoichiometry 3 moles of H₂ are formed from 54 grams of Al, 0.50 moles of H₂ are formed from how much mass of Al?

mass of Al= 9 grams

In summary, the correct answer is option B): 9 grams of aluminum are necessary to produce 0.50 mol of hydrogen gas at 25°C and 1.0 atm