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patriot [66]
3 years ago
14

How to do this question plz answer

Chemistry
1 answer:
Ira Lisetskai [31]3 years ago
6 0

Explanation:

This is a pretty simple question of mole concept and the solution goes like this:

Given:

Volume of NaOH = 500 mL

Molarity of the aqueous solution = 0.5 M

Molar mass of NaOH= (23+16+1)g = 40g

Let the mass of NaOH be x.

Now , according to formula number of moles of a given substance = given mass/molar mass

Therefore, no. Of moles of NaOH= x/40

Since, molarity = number of moles of solute/ volume of solution in lit.

Hence, 0.5 = (x/40)/(500/1000)

Solving this we get x= 10g

Hence, 10g of NaOH should be taken to obtain a 0.5 M solution of volume 500 mL.

I hope I could explain the numerical well and help you.

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Leftover: approximately 11.73 g of sulfuric acid.

<h3>Explanation</h3>

Which reactant is <em>in excess</em>?

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How many <em>moles</em> of H₂SO₄ is consumed?

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Each mole of Al(OH)₃ corresponds to 3/2 moles of H₂SO4. The formula 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 excess, meaning that all 0.19230 moles will be consumed. Accordingly, 0.19230 × 3/2 = 0.28845 moles of H₂SO₄ will be consumed.

How many <em>grams</em> of H₂SO₄ is consumed?

The molar mass of H₂SO₄ is 98.076 g.mol. The mass of 0.28845 moles of H₂SO₄ is 0.28845 × 98.076 = 28.289 g.

How many <em>grams</em> of H₂SO₄ is in excess?

40 grams of sulfuric acid H₂SO₄ is available. 28.289 grams is consumed. The remaining 40 - 28.289 = 11.711 g is in excess. That's closest to the first option: 11.73 g of sulfuric acid.

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