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Consider the balanced equation, and answer the questions that follow. ALOH)3 + 3 HBr -> 34,0 + Alors When 8.09 grams of HBr hydrobromic acid reacts with an excess of Al(OH)3, aluminum hydroxide... How many grams of H2O would be produced?

a) 4.53 grams
b) 5.62 grams
c) 6.78 grams
d) 8.09 grams

1 Answer

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Final answer:

To find the mass of H2O produced, convert the given mass of HBr to moles, and then use stoichiometry to determine the number of moles of H2O. Finally, multiply the number of moles of H2O by the molar mass of water to find the mass of H2O produced. 5.46 grams.

Step-by-step explanation:

The balanced equation for the reaction between Al(OH)3 and HBr is:

Al(OH)3 + 3 HBr → AlBr3 + 3 H2O

In this equation, 1 mole of Al(OH)3 reacts with 3 moles of HBr to produce 3 moles of H2O. To find the amount of H2O produced, we need to convert the given mass of HBr to moles and then use stoichiometry to determine the number of moles of H2O. Given that 8.09 grams of HBr is reacting, we can calculate the number of moles of HBr as follows:

moles of HBr = mass of HBr / molar mass of HBr

Using the molar mass of HBr = 80.91 g/mol, we get:

moles of HBr = 8.09 g / 80.91 g/mol = 0.1 mol

Since the stoichiometry of the balanced equation tells us that 1 mole of HBr produces 3 moles of H2O, we can calculate the number of moles of H2O produced:

moles of H2O = moles of HBr × (3 moles of H2O / 1 mole of HBr) = 0.1 mol × 3 = 0.3 mol

Finally, to determine the mass of H2O produced, we can use the molar mass of H2O (18.02 g/mol) and the number of moles of H2O:

mass of H2O = moles of H2O × molar mass of H2O = 0.3 mol × 18.02 g/mol ≈ 5.46 grams.

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