Question

2-Aluminum carbide, Al4C3, reacts with HCl to produce aluminum chloride and methane gas, CH4.How many liters of CH4 (g) at STP will be produced by the reaction of 0.2 mol of Al4C3 with 1.2 mol of HCl?

Answer 1

Using stoichiometric calculations, it is determined that 0.2 moles of Al4C3 reacts to produce 0.6 moles of CH4 at STP, which equates to 13.44 liters. However, since HCl is the limiting reagent, the actual yield of CH4 would be less than 13.44 liters.

Step-by-step explanation:

The student asked how many liters of methane gas (CH4) at STP would be produced from the reaction of 0.2 mol of aluminum carbide (Al4C3) with hydrochloric acid (HCl). To solve this, the balanced chemical equation needs to be written, which is:

Al4C3 + 12 HCl → 4 AlCl3 + 3 CH4

Next, we perform stoichiometric calculations to find out the amounts of reactants and products. This equation shows that 1 mole of Al4C3 reacts with 12 moles of HCl to produce 3 moles of CH4. Since the student has 0.2 moles of Al4C3, it will produce 0.2 x 3 = 0.6 moles of CH4. At STP (standard temperature and pressure), 1 mole of any gas occupies 22.4 liters. Thus, the volume of CH4 produced is 0.6 moles x 22.4 L/mole = 13.44 liters.

However, because only 1.2 moles of HCl are available and 12 moles are needed for complete reaction with 1 mole of Al4C3, HCl is the limiting reagent in this scenario. Therefore, HCl will completely react and produce less than 13.44 liters of CH4.