Question

2AI + 6HCI -> 2AICI3+ 3H2 aluminum reacts with hydrochloric acid. How many moles of aluminum are necessary to produce 11 L of hydrogen gas at STP?

Answer 1

Answer:

0.33 moles of aluminum are necessary to produce 11 L of hydrogen gas at STP.

Explanation:

The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases. And in these conditions 1 mole of any gas occupies an approximate volume of 22.4 liters.

You can apply the following rule of three: if by STP conditions, 22.4 L represents 1 mole of hydrogen, 11 L how many moles will they be?

$amount of moles of hydrogen=\frac{11 L*1 mole}{22.4 L}$

amount of moles of hydrogen= 0.491 moles

On the other side, the balanced reaction is:

2 AI + 6 HCI → 2 AICI₃ + 3 H₂

By reaction stoichiometry, the following amounts of moles of each compound participate in the reaction:

• AI: 2 moles
• HCI: 6 moles
• AICI₃: 2 moles
• H₂: 3 moles

Then you can apply the following rule of three: if 3 moles of hydrogen are produced by 2 moles of aluminum, 0.491 moles of hydrogen are produced by how many moles of aluminum?

$amount of moles of aluminum=\frac{0.491 moles of hydrogen*2 moles of aluminum}{3 moles of hydrogen}$

amount of moles of aluminum= 0.33 moles

0.33 moles of aluminum are necessary to produce 11 L of hydrogen gas at STP.