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kvv77 [185]
3 years ago
10

The following reaction shows the products when sulfuric acid and aluminum hydroxide react.

Chemistry
1 answer:
Elden [556K]3 years ago
6 0

The correct answer is approximately 11.73 grams of sulfuric acid.

The theoretical yield of water from Al(OH)3 is lower than that of H₂SO₄. As a consequence, Al(OH)3 is the limiting reactant, H₂SO₄ is in excess.

The balanced equation is:

2Al(OH)₃ + 3H₂SO₄ ⇒ Al₂(SO₄)₃ + 6H₂O

Each mole of Al(OH)3 corresponds to 3/2 moles of H₂SO₄. The molecular mass of Al(OH)3 is 78.003 g/mol. There are 15/78.003 = 0.19230 moles of Al(OH)3 in the five grams of Al(OH)3 available. Al(OH)3 is in limiting, which means that all 0.19230 moles will be consumed. Accordingly, 0.19230 × 3/2 = 0.28845 moles of H₂SO₄ will be consumed.

The molar mass of H₂SO₄ is 98.706 g/mol. The mass of 0.28845 moles of H₂SO₄ is 0.28845 × 98.706 = 28.289 g

40 grams of sulfuric acid is available, out of which 28.289 grams is consumed. The remaining 40-28.289 = 11.711 g is in excess, which is closest to the first option, that is, 11.73 grams of H₂SO₄.

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Answer:

The answer to your question is 0.41 moles

Explanation:

Data

moles of NaCl = ?

mass of NaCl = 24 g

Process

To solve this problem just calculate the molar mass of NaCl, and remember that the molar mass of any substance equals to 1 mol.

1.- Calculate the molar mass

NaCl = 23 + 35.5 = 58.5 g

2.- Use proportions and cross multiplication

               58.5 g of NaCl ------------------- 1 mol

               24.0 g               ------------------- x

                     x = (24 x 1) / 58.5

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Find the volume of a gas at standard pressure if its volume at 1.9 atm is 80 ml?
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Answer:

1.5 × 10² mL

Explanation:

Step 1: Given data

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Step 2: Calculate the final volume of the gas

For an ideal gas, we can calculate the final volume of the gas using Boyle's law.

P₁ × V₁ = P₂ × V₂

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Answer:

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Explanation:

Given:

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The rate of change in the amount of salt is given by,

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Since tank is initially filled with water so we write that,

x(0) = 0

Let amount of salt in the solution is c,

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Therefore, the rate of change in the amount of salt is \frac{dx}{dt} =( 5c}{ - \frac{x }{20})

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