Question

questions 4 through 8 use the following reaction between Aluminum and hydrochloric acid 2 Al(s) + 6 HCI (aq) → 2 AICI: (aq) + 3 H2(g) When 3.95 moles aluminum reacts with excess hydrochloric acid how many moles of hydrogen gas form? When 10.6 moles hydrochloric acid reacts with excess aluminum how many moles of hydrogen gas form? When 3.95 moles aluminum reacts with 10.6 moles hydrochloric acid how many moles of hydrogen gas form? What is the theoretical yield of hydrogen in grams for the above reaction? What was the limiting reagent in the above reaction? • End of questions using the same reaction.

Answer 1

Ans 1.

As according to the given chemical equation:

2 mole of Aluminum reacts with Hydrochloric acid to produce = 3 moles of Hydrogen gas.

1 mole of Aluminum reacts with Hydrochloric acid to produce = 3/2 moles of Hydrogen gas.

Now, 3.95moles of Aluminum reacts with an excess of Hydrochloric acid to produce = (3/2 * 3.95) moles of Hydrogen gas.

Hence, the amount of Hydrogen gas produced by using 3.95 moles of Aluminum = 5.925 moles

Ans 2.

As according to the given chemical equation:

6 moles of HCl react with Aluminium to produce = 3 moles of Hydrogen gas.

so, 1 mole of HCl react with Aluminium to produce = 3/6 moles of Hydrogen gas

so, now 10.6 moles of HCl react with Aluminium to produce = (3/6 * 10.6) moles of Hydrogen gas

Hence, the amount of Hydrogen gas produced by using 10.6 moles of HCl = 5.3 moles

Ans 3.

As we can see from the chemical reaction given in the question:

2 mole of Aluminum reacts with 6 moles of Hydrochloric acid to produce = 3 moles of Hydrogen gas.

1 mole of Aluminum reacts with 3moles Hydrochloric acid to produce = 3/2 moles of Hydrogen gas.

so, 3.95 of Aluminium will react with = (3 * 3.95) moles of HCl

= 11.85 moles of HCl

hence, for the complete consumption of Al in the reaction, we need 11.85 moles of HCl, but we have only 10.6 moles of HCl, hence, it is the limiting reagent, which will help us to predict the amount of Hydrogen produced.

As, the 6 moles of HCl react with Aluminium to produce = 3 moles of Hydrogen gas.

so, 1 mole of HCl react with Aluminium to produce = 3/6 moles of Hydrogen gas

so, now 10.6 moles of HCl react with Aluminium to produce = (3/6 * 10.6) moles of Hydrogen gas

Hence, the amount of Hydrogen gas produced by using 10.6 moles of HCl = 5.3 moles

Ans: 4

we know that:

Theoretical yield = (amount of Hydrogen produced experimentally /amount of hydrogen expected to produced)*100

if the 3.95 moles of Al reacts to the excess of HCl than

= the amount of Hydrogen gas produced is 5.925moles (shown in above solution in ans:2)

but as, the HCl is limiting reagent as it is not enough to react with 3.95 moles of Al completely, so the amount of Hydrogen produced will depend upon the amount of HCl, i.e 10.6 moles of HCl.

So, the amount of Hydrogen gas produced = 5.3 moles (shown in above solution in ans:3)

Now, Theoretical yield = ( 5.3 moles / 5.925 ) * 100

= 89.45 %

Ans : 5

HCl is here limiting agent, as it is less than the amount of HCl required for the amount of Al taken.

Hope this helps!