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Given:2 Al(OH)3 + 3 H2SO4 ⟶ Al2(SO4)3 + 6 H2OHow many grams of aluminum sulfate will be produced if 2,711 grams of aluminum hydroxide are available and fully consumed?Write your final answer in two decimal places.Use the following mass numbers:Al - 27O - 16H - 1S - 32

Question

Given:2 Al(OH)3 + 3 H2SO4 ⟶ Al2(SO4)3 + 6 H2OHow many grams of aluminum sulfate will be produced if 2,711 grams of aluminum hydroxide are available and fully consumed?Write your final answer in two decimal places.Use the following mass numbers:Al - 27O - 16H - 1S - 32

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Solution

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

  1. Calculate the molar mass of Al(OH)3 and Al2(SO4)3.

    The molar mass of Al(OH)3 = 227 (Al) + 61 (H) + 316 (O) = 78 g/mol The molar mass of Al2(SO4)3 = 227 (Al) + 3416 (O) + 3*32 (S) = 342 g/mol

  2. Calculate the number of moles of Al(OH)3.

    Moles of Al(OH)3 = mass / molar mass = 2711 g / 78 g/mol = 34.76 moles

  3. Use the stoichiometry of the reaction to find the number of moles of Al2(SO4)3 produced.

    From the balanced chemical equation, 2 moles of Al(OH)3 produce 1 mole of Al2(SO4)3. So, 34.76 moles of Al(OH)3 will produce 34.76/2 = 17.38 moles of Al2(SO4)3.

  4. Convert the moles of Al2(SO4)3 to grams.

    Mass of Al2(SO4)3 = moles * molar mass = 17.38 moles * 342 g/mol = 5947.96 g

So, if 2,711 grams of aluminum hydroxide are fully consumed, 5947.96 grams of aluminum sulfate will be produced. Rounded to two decimal places, the answer is 5947.96 grams.

This problem has been solved

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