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F 10.90 grams of Hydrogen Bromide, HBr (molar mass= 80.91 g/mol) reacts with excess aluminum

Hydroxide, Al(OH),. How many grams of Aluminum Bromide (molar mass- 266.69 g/mol) are formed?
(Hint: Use the BALANCED chemical formula above)
Mole Ratio=
Mass=
Molar Mass=
Mass=
Molar Mass=

User Yarh
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The balanced chemical equation for the reaction between hydrogen bromide (HBr) and aluminum hydroxide (Al(OH)3) is:

3 HBr + Al(OH)3 → AlBr3 + 3 H2O

The mole ratio between HBr and AlBr3 is 3:1. This means that for every 3 moles of HBr, 1 mole of AlBr3 is produced.

Now let's calculate the answers to the questions:

Mole Ratio:
The mole ratio of HBr to AlBr3 is 3:1. This means that for every 3 moles of HBr, we get 1 mole of AlBr3.

Mass:
The given mass of HBr is 10.90 grams.

Molar Mass:
The molar mass of HBr is 80.91 g/mol, and the molar mass of AlBr3 is 266.69 g/mol.

Mass:
To calculate the mass of aluminum bromide formed, we need to determine the number of moles of HBr and then use the mole ratio to find the number of moles of AlBr3. Finally, we can multiply the moles of AlBr3 by its molar mass to get the mass.

Let's calculate the values:

Number of moles of HBr = mass of HBr / molar mass of HBr
Number of moles of HBr = 10.90 g / 80.91 g/mol ≈ 0.1347 mol

Number of moles of AlBr3 = (1/3) * number of moles of HBr
Number of moles of AlBr3 = (1/3) * 0.1347 mol ≈ 0.0449 mol

Mass of AlBr3 = number of moles of AlBr3 * molar mass of AlBr3
Mass of AlBr3 = 0.0449 mol * 266.69 g/mol ≈ 11.98 g

Therefore, approximately 11.98 grams of aluminum bromide are formed in the reaction.
User Arie
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