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son4ous [18]
2 years ago
8

How many grams of aluminum oxide will you need to start with in order to make 6500 g of aluminum hydroxide?

Chemistry
2 answers:
REY [17]2 years ago
8 0

Answer:

56.7 grams of aluminum oxide (AI²O³) are produced.Need a fast expert's response.

kherson [118]2 years ago
3 0

Answer:

Explanation:

2Al(s) + 3 2 O2(g) → Al2O3(s) And given the stoichiometry ...and EXCESS dioxygen gas...we would get 6.25⋅ mol of alumina. the which represents a mass... ...6.25 ⋅ mol ×101.96 ⋅ g ⋅ mol−1 molar mass of alumina ≡ 637.25 ⋅ g.

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The concentrated sulfuric acid we use in the laboratory is 98.0% sulfuric acid by weight. Calculate the molality and molarity of
timama [110]

Answer : The molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.

Solution : Given,

Density of solution = 1.83g/cm^3=1.83g/ml

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98.0 % sulfuric acid by mass means that 98.0 gram of sulfuric acid is present in 100 g of solution.

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Mass of solution = 100 g

Mass of solvent = Mass of solution - Mass of solute = 100 - 98.0 = 2 g

First we have to calculate the volume of solution.

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Now we have to calculate the molarity of solution.

Molarity=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{volume of solution}}=\frac{98.0g\times 1000}{98.079g/mole\times 54.64ml}=18.29mole/L

Now we have to calculate the molality of the solution.

Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}=\frac{98.0g\times 1000}{98.079g/mole\times 2g}=499.59mole/Kg

Therefore, the molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.

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