Question

Give the theoretical yield, in moles, of H2 from the reaction of 3.8 moles of Al with 4.0 L of 2,5 M solution of HCI2Al(s) + 6HCl(aq) ---> 2A1C13(aq) + 3H2(8)O 5.7 moles of H2, Al being the LRO 5.7 moles of H2, HCl being the LR5.0 moles of H2, Al being the LRO 5.0 moles of H2, HCl being the LRA Moving to another question will save this response.

Answer 1

We have the balanced equation for the reaction.

`2Al_+6HCl\rightarrow2AlCl_3+3H_2`

To determine which is the limiting reactant we must know the moles of both reactants. We have the moles of Al equal to 3.8 mol. We calculate the moles of HCl with the data that they give us of molarity and volume in the following way:

`molHCl=2.5(mol)/(L)*4.0L=10molHCl`

Now, the ratio according to the HCl to Al reaction equation is 6/2=3/1.

The HCl to Al ratio that we have according to the available moles is: 10/3.8=2.63. This ratio is lower than the theoretical one, that is, there are not enough moles of HCl if all the moles of Al are to be reacted. Therefore, HCl is the limiting reactant.

The calculations are made according to the moles of HCl available. Now the ratio H2 to HCl is 3/6=1/2. So the moles of H2 will be:

`\begin{gathered} molH_2=GivenmolHCl*(1molH_2)/(2molHCl) \\ molH_2=10molHCl*(1molH_(2))/(2molHCl)=5.0molH_2 \end{gathered}`

The theoretical yield is 5.0molH2.

The answer is: 5.0 moles of H2, HCl being the LR. 4th option