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Ede4ka [16]
2 years ago
9

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.
Chemistry
1 answer:
Ira Lisetskai [31]2 years ago
6 0

Answer:

a. 167 mL.

b. 39.3 %.

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*\frac{1molAl}{27.0gAl} *\frac{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=\frac{n}{M}=\frac{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*\frac{1molAl_2(SO_4)_3}{3molH_2SO_4}*\frac{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=\frac{112g}{285.13g}*100\%\\\\Y=39.3\%

Best regards!

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