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1. If I combine 20.0 g of aluminum oxide with excess water, it forms aluminum hydroxide.

a) How many grams of aluminum hydroxide are formed?

Sagot :

GinnyAnsweridn
Sure, let's solve the problem step-by-step.

Step 1: Identify the given data.

- The molar mass of Aluminum Oxide ([tex]\( \text{Al}_2\text{O}_3 \)[/tex]) is 101.96 g/mol.
- The molar mass of Aluminum Hydroxide ([tex]\( \text{Al(OH)}_3 \)[/tex]) is 78.00 g/mol.
- The given mass of Aluminum Oxide is 20.0 grams.

Step 2: Calculate the number of moles of Aluminum Oxide ([tex]\( \text{Al}_2\text{O}_3 \)[/tex]).

[tex]\[ \text{Moles of } \text{Al}_2\text{O}_3 = \frac{\text{Given mass of } \text{Al}_2\text{O}_3}{\text{Molar mass of } \text{Al}_2\text{O}_3} \][/tex]

[tex]\[ \text{Moles of } \text{Al}_2\text{O}_3 = \frac{20.0 \, \text{grams}}{101.96 \, \text{g/mol}} \][/tex]

[tex]\[ \text{Moles of } \text{Al}_2\text{O}_3 = 0.19615535504119264 \, \text{moles} \][/tex]

Step 3: Determine the stoichiometric relationship between Aluminum Oxide ([tex]\( \text{Al}_2\text{O}_3 \)[/tex]) and Aluminum Hydroxide ([tex]\( \text{Al(OH)}_3 \)[/tex]).

According to the balanced chemical equation:

[tex]\[ \text{Al}_2\text{O}_3 + 3 \, \text{H}_2\text{O} \rightarrow 2 \, \text{Al(OH)}_3 \][/tex]

From the equation, 1 mole of [tex]\( \text{Al}_2\text{O}_3 \)[/tex] produces 2 moles of [tex]\( \text{Al(OH)}_3 \)[/tex].

Step 4: Calculate the number of moles of Aluminum Hydroxide ([tex]\( \text{Al(OH)}_3 \)[/tex]).

[tex]\[ \text{Moles of } \text{Al(OH)}_3 = 2 \times \text{Moles of } \text{Al}_2\text{O}_3 \][/tex]

[tex]\[ \text{Moles of } \text{Al(OH)}_3 = 2 \times 0.19615535504119264 \, \text{moles} \][/tex]

[tex]\[ \text{Moles of } \text{Al(OH)}_3 = 0.39231071008238527 \, \text{moles} \][/tex]

Step 5: Calculate the mass of Aluminum Hydroxide ([tex]\( \text{Al(OH)}_3 \)[/tex]).

[tex]\[ \text{Mass of } \text{Al(OH)}_3 = \text{Moles of } \text{Al(OH)}_3 \times \text{Molar mass of } \text{Al(OH)}_3 \][/tex]

[tex]\[ \text{Mass of } \text{Al(OH)}_3 = 0.39231071008238527 \, \text{moles} \times 78.00 \, \text{g/mol} \][/tex]

[tex]\[ \text{Mass of } \text{Al(OH)}_3 = 30.60023538642605 \, \text{grams} \][/tex]

Final Answer:

a) The number of grams of aluminum hydroxide ([tex]\( \text{Al(OH)}_3 \)[/tex]) formed is 30.60023538642605 grams.
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