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Montano1993 [528]
3 years ago
13

A 3.00 g sample of unknown hydrocarbon is prepared for combustion analysis. after the hydrocarbon undergoes complete combustion,

8.80 g of co2 and 5.40 g of h2o are produced. what is the empirical formula of the unknown?
Chemistry
1 answer:
zmey [24]3 years ago
7 0

Answer:

CH₃.

Explanation:

  • The complete combustion of hydrocarbon in an excess of oxygen produces CO₂ and H₂O according to the general reaction equation:

CₐHₓ + O₂(excess) → b CO₂ + c H₂O,

Where a is the no. of C atoms in the unknown hydrocarbon.

x is the no. of H atoms in the unknown hydrocarbon.

b is the no. of moles of CO₂ produced from the combustion of unknown hydrocarbon.

c is no. of moles of H₂O produced from the combustion of unknown hydrocarbon.

  • We can calculate b and c using the relation:

n = mass / molar mass.

  • n of CO₂ = (8.80 g) / (44.0 g/mol) = 0.2 mol.

Every molecule of CO₂ contains 1 C atom, so no. of moles of C atoms in CO₂ and thus in the original hydrocarbon is 0.2 mol.

  • n of H₂O = (5.40 g) / (18.0 g/mol) = 0.3 mol.

Every molecule of H₂O contains 2 H atoms, so no. of moles of H atoms in H₂O and thus in the original hydrocarbon is (2 x 0.3) 0.6 mol.

  • The ratio of C to H in the unknown hydrocarbon is (0.2): (2 x 0.3) = 0.2: 0.6.
  • Dividing by the lowest no. of moles (0.2) that of C.

∴ The ratio of C to H in the unknown hydrocarbon is 1: 3.

<em>So, the empirical formula of the unknown hydrocarbon is CH₃.</em>

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Explanation:

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On the other hand, sodium acetate (CH_{3}COONa) is a salt of weak acid, that is, CH_{3}COOH and strong base, that is, NaOH. Therefore, aqueous solution of sodium acetate will be basic in nature.

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When sodium acetate is mixed into this solution then it will dissociate as follows.

            CH_{3}COO^{-}Na^{+}(aq) \rightleftharpoons CH_{3}COO^{-}(aq) + Na^{+}(aq)

As both solutions are liberating acetate ion upon dissociation. Hence, it is the common ion.

So, when more acetate ions will increase from dissociation of sodium acetate the according to Le Chatelier's principle the equilibrium will shift on left side.

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Bromophenol blue is the indicator used in detecting the endpoint for the antacid analysis in this experiment. what is the expect
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snow_lady [41]

Answer:

1. Mass of KCl produced = 774.8 g of KCl

2. Mass of KNO₃ produced = 13.837g

3. Moles of NaOH made = 0.846 moles

4. Moles of LiCl produced = 0.846 moles

5. Moles of CO₂ produced = 207.6 moles

Explanation:

1. From the equation of reaction, 1 mole of ZnCl₂ produces, 2 moles of KCl.

5.02 moles of ZnCl₂ will produce, 2 × 5.02 moles of KCl = 10.4 moles of KCl

Molar mass of KCl = (39 + 35.5) g/mol = 74.5 g/mol

10.4 moles of KCl = 10.4 × 74.5 g

Mass of KCl produced = 774.8 g of KCl

2. Mole ratio of KNO₃ and KOH = 1:1

O.137 moles of KOH will produce 0.137 moles of KNO₃

Molar mass of KNO₃ = 101 g/mol

Mass of KNO₃ produced = 0.137 × 101 g = 13.837g

3. Molar mas of Ca(OH)₂ = 74.0 g

Moles of Ca(OH)₂ in 31.3 g = 31.3/74.0 = 0.423 moles of Ca(OH)₂

Mole ratio of NaOH and Ca(OH)₂ in the reaction = 2 : 1

Moles of NaOH made = 2 × 0.423 = 0.846 moles

4. Molar mass of MgCl₂ = 95.0 g

Moles of MgCl₂ in 40.2 g = 40.2/95.0 = 0.423 moles

From the reaction equation, mole ratio of MgCl₂ and LiCl = 1:2

Moles of LiCl produced = 2 × 0.423 = 0.846 moles

5. From the equation of reaction, 1 mole of C₆H₁₀O₅ produces 6 moles of cO₂

34.6 moles of C₆H₁₀O₅ will produce 34.6 × 6 moles of CO₂

Moles of CO₂ produced = 207.6 moles

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