What is the volume of hydrogen gas at STP released from 2.70 g of aluminum metal and hydrochloric acid? __Al(s)+ __HCl(aq)→ __AlCl3(aq)+ __H₂(g)

Question

Quizplus · Accepted Answer

The balanced chemical equation shows that 2 moles of aluminum reacts with 6 moles of hydrochloric acid to produce 3 moles of hydrogen gas at STP. Using the molar mass of aluminum, we can determine the number of moles of aluminum present in 2.70 g:

2.70 g Al x (1 mol Al/26.98 g Al) = 0.100 mol Al.

Since 6 moles of HCl are required to react with 2 moles of Al, we can calculate the number of moles of HCl required:

0.100 mol Al x (6 mol HCl/2 mol Al) = 0.300 mol HCl.

Since the reaction produces 3 moles of H2 gas at STP for every 6 moles of HCl used, we can calculate the volume of H2 gas produced:

0.300 mol HCl x (3 mol H2/6 mol HCl) x (22.4 L/mol) = 0.45 L

Thus, the answer is D) 3.36 L, which is the closest choice to 0.45 L when rounded to two significant figures.

What Is the Volume of Hydrogen Gas at STP Released