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Georgia [21]
3 years ago
12

Which equation represents a neutralization reaction? A. 4Fe(s) + 3O2(g) → 2Fe2O3(s) B. 2H2(g) + O2(g) → 2H2O() C. HNO3(aq) + KOH

(aq) → KNO3(aq) + H2O() D. AgNO3(aq) + KCl(aq) → KNO3(aq) + AgCl(s)
Chemistry
2 answers:
devlian [24]3 years ago
4 0

Answer : The correct option is, (C)

Explanation :

A. 4Fe(s)+3O_2(g)\rightarrow 2Fe_2O_3(s)

This is synthesis reaction in which the two or more reactant react to give a single product.

B. 2H_2(g)+O_2(g)\rightarrow 2H_2O(l)

This is combination reaction in which the two reactants combine to form a single product.

C. HNO_3(aq)+KOH(aq)\rightarrow KNO_3(aq)+H_2O(l)

This is a neutralization reaction in which an acid react with a base to give a salt and water as a product.

D. AgNO_3(aq)+KCl(aq)\rightarrow KNO_3(aq)+AgCl(s)

This is a double displacement reaction in which the positive cation and negative anion of two reactants exchange their places to give two new products.

Hence, the reaction (C) is a neutralization reaction.

stepladder [879]3 years ago
3 0
The equation represents a neutralization reaction is HNO3(aq) + KOH(aq) → KNO3(aq) + H2O(). The answer is letter C. A. 4Fe(s) + 3O2(g) → 2Fe2O3(s), 2H2(g) + O2(g) → 2H2O() and <span>AgNO3(aq) + KCl(aq) → KNO3(aq) + AgCl(s) are not neutralization reaction.</span>
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When 0.300 g of a diprotic acid was titrated with 0.100 M LiOH, 40.0 mL of the LiOH solution was needed to reach the second equi
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Answer:

  • <em><u>H₂C₄H₄O₆</u></em>

Explanation:

The clue of this question is to find the molar mass of the <em>diprotic acid</em> and compare witht the molars masses of the choices' acid to identify the formula of the diprotic acid.

The procedure is:

  1. Find the number of moles of the base: LiOH
  2. Use stoichionetry to infere the number of moles of the acid.
  3. Use the formula molar mass = mass in grams / number of moles, to find the molar mass of the diprotic acid.
  4. Compare and conclude.

<u>Solution:</u>

<u>1. Number of moles of the base, LiOH:</u>

  • M = n / V in liter ⇒ n = M × V = 0.100 M × 40.0 ml × 1 liter / 1,000 ml = 0.004 mol LiOH.

<u>2. Stoichiometry:</u>

Since this a neutralization reaction of a diprotic acid with a mono hydroxide base (LiOH), the mole ratio at the second equivalence point is: 2 mol of base / 1 mole of acid; because each mole of LiOH releases 1 mol of OH⁻, while each mole of diprotic acid releases 2 mol of H⁺.

Hence, 0.004 mol LiOH × 1 mol acid / 2 mol LiOH = 0.002 mol acid.

<u>3. Molar mass of the acid:</u>

  • molar mass = mass in grams / number of moles = 0.300 g / 0.002 mol = 150. g/mol

<u>4. Molar mass of the possible diprotic acids:</u>

a. H₂Se: 2×1.008 g/mol + 78.96  g/mol = 80.976 g/mol

b. H₂Te: 2×1.008 g/mol + 127.6  g/mol = 129.616 g/mol

c. H₂C₂O₄ ≈ 2×1.008 g/mol + 2×12.011 g/mol + 4×15.999 g/mol ≈ 90.034 g/mol

d. H₂C₄H₄O₆ = 6×1.008 g/mol +  4×12.011 g/mol + 6×15.999 g/mol = 150.086 g/mol ≈ 150 g/mol.

<u>Conclusion:</u> since the molar mass of H₂C₄H₄O₆ acid is 150 g/mol, you conclude that is the diprotic acid whose 0.300 g were titrated with 40.0 ml of 0.100 M LiOH solution.

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