Question

2Al(s) + 3H2SO4(aq) → Al2(SO4)3(aq) + 3H2(g)a. Determine the volume (mL) of 15.0 M sulfuric acid needed to react with 45.0 g of aluminum to produce aluminum sulfate.b. Determine the % yield if 112 g of aluminum sulfate is produced under the above conditions.

Answer 1

a. 167 mL.

b. 39.3 %.

Step-by-step explanation:

Hello!

In this case, for the undergoing chemical reaction, since 45.0 g of aluminum react, based on the 2:3 mole ratio with sulfuric acid, we can compute the required moles as shown below:

`n_(H_2SO_4)=45.0gAl*(1molAl)/(27.0gAl) *(3molH_2SO_4)/(2molAl) =2.50molH_2SO_4`

Next, since the molarity of a solution is computed based on the moles and volume (M=n/V), we can compute the required volume of sulfuric acid as shown below:

`V=(n)/(M)=(2.50mol)/(15.0mol/L)=0.167L`

That in mL is 167 mL.

Moreover, for the percent yield, we compute the grams of aluminum sulfate that are produced based on the required 2.50 moles of sulfuric acid:

`m_(Al_2(SO_4)_3)=2.50molH_2SO_4*(1molAl_2(SO_4)_3)/(3molH_2SO_4)*(342.15gAl_2(SO_4)_3)/(1molAl_2(SO_4)_3) \\\\m_(Al_2(SO_4)_3)=285.13gAl_2(SO_4)_3`

Therefore the percent yield is:

`Y=(112g)/(285.13g)*100\%\\\\Y=39.3\%`

Best regards!