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Calculate the mass of hydrogen formed when 81 g of aluminum reacts with excess hydrochloric acid according to the balanced equation below.Al + HCl → AlCl3 + H2

Question

Calculate the mass of hydrogen formed when 81 g of aluminum reacts with excess hydrochloric acid according to the balanced equation below.

Al+HClAlCl3+H2 \text{Al} + \text{HCl} \rightarrow \text{AlCl}_3 + \text{H}_2

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Solution

To solve this problem, we need to follow these steps:

  1. Write down the balanced chemical equation: 2Al + 6HCl → 2AlCl3 + 3H2

  2. Calculate the molar mass of Aluminum (Al) which is approximately 26.98 g/mol.

  3. Calculate the number of moles of Aluminum (Al) in 81g. Moles = mass/molar mass = 81g / 26.98 g/mol = 3 moles

  4. From the balanced chemical equation, we can see that the ratio of Al to H2 is 2:3. This means that for every 2 moles of Al, 3 moles of H2 are produced.

  5. So, if we have 3 moles of Al, we will have (3/2)*3 = 4.5 moles of H2.

  6. The molar mass of H2 (Hydrogen gas) is approximately 2.02 g/mol.

  7. Finally, calculate the mass of H2 formed by multiplying the number of moles by the molar mass. Mass = moles * molar mass = 4.5 moles * 2.02 g/mol = 9.09 g

So, the mass of hydrogen formed when 81 g of aluminum reacts with excess hydrochloric acid is approximately 9.09 g.

This problem has been solved

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