Question

Sulfuric acid dissolves aluminum metal according to the following reaction:
2Al (s) + 3H2SO4 (aq) --> Al2(SO4)3 (aq) + 3H2 (g)
Suppose you wanted to dissolve an aluminum block with a mass of 15.0 g.

What minimum mass of H2SO4 (in g) would you need?

What mass of H2 gas (in g) would be produced by the complete reaction of the aluminum block?

Answer 1

Answer:

$m_{H_2SO_4}=81.7gH_2SO_4$

$m_{H_2}=1.67gH_2$

Explanation:

Hello,

Based on the given undergoing chemical reaction is is rewritten below:

$2Al (s) + 3H_2SO_4 (aq)\rightarrow Al _2(SO4)_3 (aq) + 3H_2 (g)$

By stoichiometry we find the minimum mass of H2SO4 (in g) as shown below:

$m_{H_2SO_4}=15.0gAl*\frac{1molAl}{27gAl}*\frac{3molH_2SO_4}{2molAl}*\frac{98gH_2SO_4}{1molH_2SO_4} \\m_{H_2SO_4}=81.7gH_2SO_4$

Moreover, mass of H2 gas (in g) would be produced by the complete reaction of the aluminum block turns out:

$m_{H_2}=15.0gAl*\frac{1molAl}{27gAl}*\frac{3molH_2}{2molAl}*\frac{2gH_2}{1molH_2} \\m_{H_2}=1.67gH_2$

Best regards.

Answer 2

Answer:

First part: m = 78.5  g

Second part: m = 1.61 g

Explanation:

First part:

The molar mass of aluminum is used to determine the number of moles of aluminum. Moles of aluminum is converted to moles of  H2SO4  using the molar ratio and then to the mass of  H2SO4  using the molar mass of  H2SO4.

Second part:

Dividing by the molar mass of aluminum gives the number of moles of aluminum, which can be converted to moles of  H2  using the molar ratio. Multiplying by the molar mass of  H2  gives  g  H2 .