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.

a. 167 mL.

b. 39.3 %.

Explanation:

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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:

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:

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:

Therefore the percent yield is:

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- aluminum and copper chloride.

- the chloride ion to separate the aluminum from the oxygen so that the aluminum can react with the copper ions (and the water molecules).